Question

6. A baseball player hits a ball from 1.3 m above the ground with an initial velocity of 36 m/s at an angle of 45. Will it clear a 7.5 m wall located 125 m away?

Answer 1

Answer on Question #42939 - Physics - Mechanics | Kinematics | Dynamics

6. A baseball player hits a ball from $1.3\mathrm{m}$ above the ground with an initial velocity of $36\mathrm{m/s}$ at an angle of 45 deg. Will it clear a $7.5\mathrm{m}$ wall located $125\mathrm{m}$ away?

Solution:

$S = 125\mathrm{m}$ - horizontal range of the wall;

$V = 36\frac{m}{s}$ - initial speed of the ball;

$\alpha = 45{}^{\circ}$ - the angle between the velocity and the horizontal;

$\mathrm{H} = 10\mathrm{m}$ - initial height;

$\mathrm{h} = 7.5\mathrm{m}$ - height of the wall;

Equation of the motion for the ball, directed at angle $\alpha$ : (t - time of the flight)

$V_{x} = V\cos \alpha ;V_{y} = V\sin \alpha ;$

x: $S = V_{x}t = Vt\cos \alpha$

$$t = \frac {S}{V \cos \alpha}$$

If the height of the ball $(\mathrm{h_b})$ will be more than $\mathrm{H}$ when it's distance will be $S$ , ball will clear the wall away:

y: $h_b - H = Vt \sin \alpha - \frac{gt^2}{2}$

$$\mathrm{h}_{\mathrm{b}} = \mathrm{H} + \mathrm{Vt} \sin \alpha - \frac{\mathrm{g} \mathrm{t}^{2}}{2}$$

(1) in(2):

$$\begin{aligned} \mathrm{h}_{\mathrm{b}} = \mathrm{H} + \mathrm{V} \frac{\mathrm{S}}{\mathrm{V} \cos \alpha} \sin \alpha - \frac{\mathrm{g} \left(\frac{\mathrm{S}}{\mathrm{V} \cos \alpha}\right)^{2}}{2} &= \mathrm{H} + \mathrm{S} \tan \alpha - \frac{\mathrm{g}}{2} \left(\frac{\mathrm{S}}{\mathrm{V} \cos \alpha}\right)^{2} \\ &= 10 \mathrm{m} + 125 \mathrm{m} \cdot \tan 45{}^{\circ} - \frac{9.8 \frac{\mathrm{m}}{\mathrm{s}^{2}}}{2} \left(\frac{125 \mathrm{m}}{36 \frac{\mathrm{m}}{\mathrm{s}} \cos 45{}^{\circ}}\right)^{2} = 16.85 \mathrm{m} \end{aligned}$$

$\mathrm{h}_{\mathrm{b}} > \mathrm{h}$ because $16.85 \mathrm{m} > 7.5 \mathrm{m} \Rightarrow$ ball will clear the wall away:

**Answer**: ball will clear the wall away:

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