Question

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?

Accepted Answer

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

Answer

@macbeth9
 
a problem can have more than one method of solving it but cannot have more than one answers !
 
Let me tell you where you have gone wrong:
 
The question was not about what is the height maximum reached by the ball when he had thrown it at 45 deg and achieved 100 m range ! The question was about the ability of the cricketer to throw the ball up !
 
So its not a single throw. Let me explain you:
 
1) Suppose you have seen that cricketer throwing that ball throwing 100 m away manytimes with his full force, and you have calculated using the maximum range formula that the cricketer has that much power in his arms. That is his maximum throwing speed (say V_max).
 
2) After knowing the power of throwing the ball of the cricketer you have estimated using a different formula with speed v_max angle=/=45 deg but 90 deg because now he will be throwing straight up of course and this is the height that was asked in the question. 
 
Now next time tell that cricketer to throw the ball up, you will see that now the answer is
 
50 m and not 25 m.

Answer

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.

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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?

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