Question

A student is able to throw a ball vertically to maximum height of $40m$. The maximum distance to which the student can throw the ball in the horizontal direction:

• A. $40{\left(2\right)}^{1/2}m$
• B. $20{\left(2\right)}^{1/2}m$
• C. $20m$
• D. $80m$

Answer 1

The correct option is D $80m$

Let the maximum velocity with which the student can throw be $v$.

Thus $0^2-v^2=2\times -g\times s$

$⟹v^2=800m^2/s^2$

Now for maximum range to be obtained, the projectile must be thrown at angle of ${45}^{\circ }$ from the horizontal.

Thus the range of the projectile $=$ $\left(vcos\theta \right)t=\frac{2v^{2}cos\theta sin\theta }{g}=80m$