Question

A football is kicked at an angle θ = ${45}^{\circ }$. It is intended that the ball lands in the back of a moving truck, which has a trunk of length L = 5 m. If the initial horizontal distance from the back of the truck to the ball, at the instant of the kick is 5 m & the truck moves directly away from the ball at velocity V = 9 m/s. What is approximate maximum & minimum velocity (in m/s) so that ball lands on the truck? (Assume the ball & back of the truck to be at the same horizontal level)

• A. (16, 17)
• B. (16, 18)
• C. (15, 17)
• D. (15, 18)

Answer 1

The correct option is A

(16, 17)

Finding minimum velocity:

In the case of minimum velocity the ball just lands on the back end of the truck which is initially 5 m away from the starting position of the ball. Suppose the ball lands after time 't'. In time t, truck would have moved the distance = (9t) m

Total distance u = (9t) m = 5 m which should be the range in order to land it successfully.

We know = $\frac{u^{2}sin2\theta }{g}=\frac{u^{2}sin{90}^o}{10}=(9t+5)$

$t=\frac{2usin\theta }{g}=\frac{2usin{45}^o}{10}$

$=\frac{\sqrt{2}}{10}u$

$\frac{u^2}{10}=\frac{9\sqrt{2}u}{10}+50$

$u^2=9\sqrt{2}u+500=0$

What is the solution?

$a{x}^2+bx+c=0$

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

{put some effort}

u = 15.82 $\frac{m}{s}$ (approx.)

For maximum velocity

For the same reason,

R = (5+5+9t)=(10+9t)=$\frac{u^{2}sin2\theta }{g}$

Where t = $\frac{2usin\theta }{g}\Rightarrow$ u max = 17.04 $\frac{m}{s}$