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

*print*Print*list*Cite

### 3 Answers

The cricketer can throw the ball to a maximum horizontal distance of 100 m. To do this the ball has to be thrown with a velocity v at an angle of elevation equal to 45 degrees.

The range of an object thrown at a velocity v with an angle of elevation 45 degrees is S = v^2/g

100 = v^2/g

=> v^2 = 100*g

In the same throw, the maximum height of the ball can be derived from the equation v^2 - u^2 = 2*g*h

It is is h = `(sqrt 980)^2/(2*9.8)` = `980/(2*9.8) = 50` m

### User Comments

First note that,

Maximum range of a projectile (ball in our case) R is given in terms of speed "v" and acceleration due to gravity "g" is given by:

R = v^2 /g

so

v^2 = R g --------(1)

Now "cricketer can throw the ball maximum R distance" means he has the power to deliver the ball with an energy of

KE = (1/2) m v^2 [here m is the mass of the ball]

= (1/2) m (R g)------(2)

[using (1) ]

so this is the maximum energy that the cricketer can supply to the ball when throwing. Now when he will through the ball vertically up, this same energy he will supply again, but this kinetic energy will exhaust when the ball reaches rest at some height say "H" so at this height the whole of the kinetic energy will be converted to potential energy "m g H", hence

m g H = KE = (1/2) R m g [using (2)]

=> H = R/2

given R = 100 m, so

H = 100/2 m = 50 m

so 50 m high the cricketer can throw up maximum.