Question

John kicks the ball, and the ball does a projectile motion with an angle of 53o to horizontal with an initial velocity of 10 m/s. What is (a) its maximum height, (b) its maximum distance, and (c) its hangtime?

Answer 1

If we analyze the system we have the following diagram, where the initial velocities are $V_{0x}=V_0\cos(\theta)$ and $V_{0y}=V_0\sin(\theta)$:

We also have the range of the trip or maximum horizontal displacement R and the maximum vertical displacement or H (both expressed in meters).

$\text {At half of the trip, velocity for the y-axis is zero:} \\V_y=V_{oy}-gt=0 \implies t= \cfrac{V_0 \sin\theta}{g} \\ \text{The time of the whole trip will be twice the amount} \\ \text{that it takes to reach H or V}_y=0: \\ t_{travel}= 2t=\cfrac{2V_0 \sin\theta}{g} \\ R=V_{0x}\cdot t_{travel}=\cfrac{2V^2_0 \sin\theta\cos\theta}{g} \\ \therefore R=\cfrac{V^2_0}{g} \sin {2\theta} \\ \text{On the other hand, the maximum height will be found} \\ \text{ when we substitute t on the equation for H:} \\ H=V_{oy}\cdot t-\frac{1}{2}gt^2= t(V_{oy}-\frac{1}{2}gt) \\ H= \bigg(\cfrac{V_0 \sin\theta}{g}\bigg) \bigg(V_0\sin {\theta}-\cfrac{g}{2} \bigg(\cfrac{V_0 \sin{\theta}}{g}\bigg) \bigg) \\ \therefore H=\cfrac{V^2_0}{2g} \sin^2 \theta$

Now we proceed to substitute the initial velocity of the projectile $V_0=10\frac{m}{s}$, the angle $\theta=53°$, and g = 9.80 m/s² to find the requested data:

$\\ (a)\,H=\cfrac{(V_0\sin \theta)^2}{2g} =\cfrac{((10\frac{m}{s})\sin {(53°)})^2}{2(9.80\frac{m}{s^2})} =3.2542\,m \\ \text{ } \\ (b)\,R=\cfrac{V^2_0}{g} \sin {2\theta}=\cfrac{(10\frac{m}{s})^2}{9.80\frac{m}{s^2}} \sin {(106°)}=9.8088\,m \\ \text{ } \\ (c)\,t_{travel} =\cfrac{2V_0 \sin\theta}{g}=\cfrac{2(10\frac{m}{s}) \sin{(53°)}}{9.80\frac{m}{s^2}}=1.6299\,s$

Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.