Detailed solutions to projectile problems are presented. These solutions may be better understood after reviewing the projectile equations.
An object is launched at a velocity of \(20 \, \text{m/s}\) in a direction making an angle of \(25^\circ\) upward with the horizontal.
a) What is the maximum height reached by the object?
b) What is the total flight time?
c) What is the horizontal range?
d) What is the magnitude of the velocity just before impact?
Solution to Problem 1:
a) The velocity and displacement components are:
\[ V_x = V_0 \cos\theta \] \[ V_y = V_0 \sin\theta - gt \] \[ x = V_0 \cos\theta \cdot t \] \[ y = V_0 \sin\theta \cdot t - \frac{1}{2}gt^2 \]
Given: \(V_0 = 20 \, \text{m/s}\), \(\theta = 25^\circ\), \(g = 9.8 \, \text{m/s}^2\).
Maximum height occurs when \(V_y = 0\):
\[ V_0 \sin\theta - gt = 0 \] \[ t = \frac{V_0 \sin\theta}{g} = \frac{20 \sin 25^\circ}{9.8} \approx 0.86 \, \text{s} \]
Maximum height:
\[ y_{\text{max}} = 20 \sin 25^\circ (0.86) - \frac{1}{2}(9.8)(0.86)^2 \approx 3.64 \, \text{m} \]
b) Flight time found from \(y=0\):
\[ V_0 \sin\theta \cdot t - \frac{1}{2}gt^2 = 0 \] \[ t\left(V_0 \sin\theta - \frac{1}{2}gt\right) = 0 \]
Solutions: \(t_1 = 0\) (launch) and \(t_2 = \frac{2V_0 \sin\theta}{g}\) (impact).
\[ \text{Time of flight} = t_2 - t_1 = \frac{2(20)\sin 25^\circ}{9.8} \approx 1.72 \, \text{s} \]
c) Horizontal range:
\[ x(t_2) = V_0 \cos\theta \cdot t_2 = \frac{2V_0^2 \cos\theta \sin\theta}{g} = \frac{V_0^2 \sin 2\theta}{g} \] \[ = \frac{20^2 \sin 50^\circ}{9.8} \approx 31.26 \, \text{m} \]
d) Velocity components at impact (\(t = t_2\)):
\[ V_x = V_0 \cos\theta = 20 \cos 25^\circ \] \[ V_y = V_0 \sin\theta - g\left(\frac{2V_0 \sin\theta}{g}\right) = -V_0 \sin 25^\circ \]
Magnitude:
\[ V = \sqrt{V_x^2 + V_y^2} = \sqrt{(20 \cos 25^\circ)^2 + (-20 \sin 25^\circ)^2} = 20 \, \text{m/s} \]
A projectile is launched from point O at \(22^\circ\) with initial velocity \(15 \, \text{m/s}\) up an incline plane that makes \(10^\circ\) with the horizontal. The projectile hits the incline at point M.
a) Find the time to hit the incline.
b) Find the distance OM.
Solution to Problem 2:
a) Displacement components:
\[ x = V_0 \cos(\theta + \alpha) \cdot t \] \[ y = V_0 \sin(\theta + \alpha) \cdot t - \frac{1}{2}gt^2 \]
where \(\theta = 22^\circ\), \(\alpha = 10^\circ\), so \(\theta + \alpha = 32^\circ\).
On the incline: \(\tan\alpha = \frac{y}{x}\). Substituting:
\[ \tan 10^\circ = \frac{V_0 \sin 32^\circ \cdot t - \frac{1}{2}gt^2}{V_0 \cos 32^\circ \cdot t} \]
Simplifying:
\[ \frac{1}{2}gt + V_0 \cos 32^\circ \tan 10^\circ - V_0 \sin 32^\circ = 0 \] \[ t = \frac{V_0 \sin 32^\circ - V_0 \cos 32^\circ \tan 10^\circ}{0.5g} \] \[ = \frac{15 \sin 32^\circ - 15 \cos 32^\circ \tan 10^\circ}{0.5(9.8)} \approx 1.16 \, \text{s} \]
b) Distance OM:
\[ OM = \sqrt{x^2 + y^2} \] \[ = \sqrt{(15 \cos 32^\circ \cdot 1.16)^2 + \left(15 \sin 32^\circ \cdot 1.16 - \frac{1}{2}(9.8)(1.16)^2\right)^2} \approx 15.0 \, \text{m} \]
A projectile is launched at \(30^\circ\) so that it falls beyond a pond of length 20 meters.
a) What is the range of initial velocities so that the projectile falls between points M and N?
Solution to Problem 3:
a) Range formula: \( R = \frac{V_0^2 \sin 2\theta}{g} \)
We need: \( OM < R < ON \) where \( OM = 30 \, \text{m} \), \( ON = 40 \, \text{m} \)
\[ 30 < \frac{V_0^2 \sin 60^\circ}{9.8} < 40 \] \[ \sqrt{\frac{30 \cdot 9.8}{\sin 60^\circ}} < V_0 < \sqrt{\frac{40 \cdot 9.8}{\sin 60^\circ}} \] \[ 18.4 \, \text{m/s} < V_0 < 21.2 \, \text{m/s} \]
A ball is kicked at \(35^\circ\) with the ground.
a) What initial velocity hits a target 30 m away at height 1.8 m?
b) What is the time to reach the target?
Solution to Problem 4:
a) From horizontal motion: \( t = \frac{30}{V_0 \cos 35^\circ} \)
Vertical motion equation:
\[ 1.8 = V_0 \sin 35^\circ \cdot t - \frac{1}{2}gt^2 \] \[ 1.8 = 30 \tan 35^\circ - \frac{1}{2}(9.8)\left(\frac{30}{V_0 \cos 35^\circ}\right)^2 \]
Solving: \[ V_0 = \sqrt{\frac{9.8 \cdot 30^2}{2 \cos^2 35^\circ (30 \tan 35^\circ - 1.8)}} \approx 18.5 \, \text{m/s} \]
b) Time: \( t = \frac{30}{18.5 \cos 35^\circ} \approx 2.0 \, \text{s} \)
A ball kicked from ground level at \(60 \, \text{m/s}\) reaches a horizontal distance of 200 m.
a) What is the launch angle \(\theta\)?
b) What is the time of flight?
Solution to Problem 5:
a) Two expressions for time of flight:
\[ T = \frac{200}{V_0 \cos\theta} = \frac{2V_0 \sin\theta}{g} \]
Equating:
\[ \frac{200}{60 \cos\theta} = \frac{2(60) \sin\theta}{9.8} \] \[ 60^2 \sin 2\theta = 200 \cdot 9.8 \] \[ \sin 2\theta = \frac{200 \cdot 9.8}{60^2} \approx 0.5444 \] \[ 2\theta \approx \sin^{-1}(0.5444) \approx 33.0^\circ \] \[ \theta \approx 16.5^\circ \]
b) Time: \( T = \frac{200}{60 \cos 16.5^\circ} \approx 3.48 \, \text{s} \)
A 600 g ball is kicked at \(35^\circ\) with initial velocity \(V_0\).
a) Find \(V_0\) if kinetic energy at maximum height is 22 J.
b) Find maximum height.
Solution to Problem 6:
a) At maximum height, \(V_y = 0\), so kinetic energy comes only from \(V_x = V_0 \cos 35^\circ\):
\[ 22 = \frac{1}{2}(0.6)(V_0 \cos 35^\circ)^2 \] \[ V_0^2 = \frac{44}{0.6 \cos^2 35^\circ} \] \[ V_0 = \sqrt{\frac{44}{0.6 \cos^2 35^\circ}} \approx 10.4 \, \text{m/s} \]
b) Initial kinetic energy:
\[ E_i = \frac{1}{2}(0.6)(10.4)^2 \approx 32.4 \, \text{J} \]
Energy difference equals potential energy gain:
\[ 32.4 - 22 = mgh \] \[ h = \frac{10.4}{0.6 \times 9.8} \approx 1.77 \, \text{m} \]
A projectile hits ground 1000 m away after 40 s.
a) Find launch angle \(\theta\).
b) Find initial velocity.
Solution to Problem 7:
a) Horizontal velocity: \( V_x = V_0 \cos\theta = \frac{1000}{40} = 25 \, \text{m/s} \)
Time of flight formula:
\[ T = \frac{2V_0 \sin\theta}{g} = 40 \] \[ V_0 \sin\theta = 20g = 196 \]
Dividing the two equations:
\[ \tan\theta = \frac{V_0 \sin\theta}{V_0 \cos\theta} = \frac{196}{25} = 7.84 \] \[ \theta = \arctan(7.84) \approx 82.7^\circ \]
b) Using Pythagorean theorem:
\[ V_0 = \sqrt{(V_0 \cos\theta)^2 + (V_0 \sin\theta)^2} = \sqrt{25^2 + 196^2} \] \[ = \sqrt{625 + 38416} = \sqrt{39041} \approx 197.6 \, \text{m/s} \]
Trajectory equation: \( y = -0.025x^2 + 0.5x \)
a) Find initial velocity and launch angle.
Solution to Problem 8:
Comparing with standard form: \( y = x\tan\theta - \frac{gx^2}{2V_0^2 \cos^2\theta} \)
From coefficients: \(\tan\theta = 0.5 \Rightarrow \theta = \arctan(0.5) \approx 26.6^\circ\)
Also: \( \frac{g}{2V_0^2 \cos^2\theta} = 0.025 \)
\[ V_0 = \sqrt{\frac{9.8}{2 \times 0.025 \times \cos^2 26.6^\circ}} \approx 15.7 \, \text{m/s} \]
Two balls (100 g and 300 g) are pushed horizontally from a 3 m high table with velocities 10 m/s and 15 m/s.
a) Time to hit ground.
b) Difference in impact points.
Solution to Problem 9:
a) Vertical motion independent of mass and horizontal velocity:
\[ -3 = -\frac{1}{2}gt^2 \] \[ t = \sqrt{\frac{6}{9.8}} \approx 0.78 \, \text{s} \]
b) Horizontal distances:
\[ x_A = 10 \times 0.78 = 7.8 \, \text{m} \] \[ x_B = 15 \times 0.78 = 11.7 \, \text{m} \]
Difference: \( |x_B - x_A| = 3.9 \, \text{m} \)