Question

A person is standing on a truck moving with a constant velocity of 14.7 m/s on a horizontal road. The man throws a ball in such a way that it returns to the truck after the truck has moved 58.8 m. Find the speed and the angle of projection (a) as seen from the truck, (b) as seen from the road.

Answer 1

(a) As seen from the truck is 19.6 m/s.

(b) As seen from the road  53.13°.

Explanation:

Step 1:

Given values V = 14.7 m/s

S = 58.8 m

where

v = velocity or speed (m/s, ft/s)

s = linear distance traveled (m, ft)

t = time (s)

$time = \frac {\text{linear distance traveled(m, ft)}}{ \text{velocity or speed (m\s, ft\s)}}$

$T=\frac{s}{v}$

$T=\frac{58.8}{14.7}$

T= 4 sec

Step 2:

(a) as seen from the truck,

Since ball returns back to truck its angle of projection is vertically upwards with respect to truck.

u=speed of projection.

Time = 4 second  

h=0

$h=u t-\frac{1}{2} g t^{2}$

$h=4 u-\frac{1}{2} 9.8 \times 4^{2}$

$h=4 u-\frac{1}{2} 9.8 \times 16$

$0=4 u-\frac{156.8}{2}$

$0=4 u-78.4$

$\begin{aligned}&u=\frac{78.4}{4}\\&u=\frac{78.4}{4}\end{aligned}$

u= 19.6  m/s

Step 3:

b) As seen from the road the speed of  

ball will be resultant of 2 speeds vertical

speed = u=19.6 m/s

Horizontal speed given by truck =14.7 m/s

$\begin{aligned}&u^{2}+v^{2}=\sqrt{19.6^{2}+14.7^{2}}\\&u^{2}+v^{2}=\sqrt{384.16+216.09}\end{aligned}$

$\begin{aligned}&u^{2}+v^{2}=\sqrt{600.25}\\&u^{2}+v^{2}=24.5 \mathrm{m} / \mathrm{s}\end{aligned}$

Thus angle seen from road:

$\begin{aligned}&=\tan ^{-1} \frac{19.6}{14.7}\\&=\tan ^{-1} 1.33\end{aligned}$

= 53.13°