Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Truck is velocity = $14.7\frac{m}{s}$

Man catches the ball after distance = $58.8m$

Time taken = $\frac{58.8}{14.7}$ = $4sec$

Ball should be in air for $4sec$ only

$\Rightarrow T=4sec$

$v_y=u$

$a_y=10$

$t=4$

$\Rightarrow 4=\frac{2u}{10}$

$\Rightarrow u=20\frac{m}{s}$

So the ball was thrown with $20\frac{m}{s}$ in perpendicular direction.

If I see from the road the ball will seem to have two velocity.

One in vertical direction $20\frac{m}{s}$ and one in horizontal direction = velocity of truck = $14.7\frac{m}{s}$

$\Rightarrow$

Velocity = $\sqrt{(20{)}^2+(14.7{)}^2}$ = $24.8$

Angle = $ta{n}^{-1}\left(\frac{20}{14.7}\right)=ta{n}^{-1}\left(1.36\right)$