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Question

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


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Solution

Step 1: Given

The distance the cricketer can throw the ball horizontally, R=100m

Step 2: Formulas used

For a projectile motion, we know that the range (horizontal distance covered) is given as,
R=V2sin2θg
where V is the initial velocity, θ is the angle at which object was thrown at and g is the acceleration due to gravity.

And the maximum height reached is given as,
H=V2sin2θ2g
where the symbols mean the same as described earlier.

Step 3: Calculating height

We know that the maximum horizontal distance is covered when θ=45°.

So we have,
R=V2sin2θg100=V2sin2×π4g45°=π4radV2g=100sinπ2=1(1)

We also have that,
H=V2sin2θ2gH=100sin2π42(1)H=50122sinπ4=12H=25m

Therefore, the maximum height the cricketer can throw the ball is 25m.


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