Question

A cricketer can throw a ball to a maximum horizontal distance of 100m. With the same speed how much high above the ground can the cricketer throw the same ball?

Answer 1

hi there

here's your answer

hope it helps

Answer 2

Given:

• horizontal distance of 100 m => R=100m

Need To Find:

• How high can he throw the ball ?

Solution

Maximum range, R = 100 m

it's possible only when , θ = 45° (refer to the last formula)

${R=\dfrac{u^2sin2\theta}{g}}$

$100=\dfrac{u^2sin2(45)}{g}$

$100=\dfrac{u^2sin90}{g}$

$\dfrac{u^2}{g}=100...(1)$

Apply formula for height ,

$H=\dfrac{u^2sin^2\theta}{g}$

$H=\dfrac{u^2(sin45)^2}{g}$

$H=\dfrac{1}{2}\times \dfrac{u^2}{g}$

$H=\dfrac{1}{2}\times 100$

$\red{→H=50 m}$

How to calculate the angle?

$\begin{gathered}\bf R=\dfrac{u^2sin2\theta}{g}\\\\\bf H= \dfrac{u^2sin^2\theta}{g}\\\\\bf t=\dfrac{2usin\theta}{g}\end{gathered}$

Maximum range is possible when angle of projection ( θ ) is 45°