Question

If a baseball player can throw a ball at 45° to a point 100m away horizontally to the initial vertical level, how high could he throw it vertically upward?

Answer 1

`50 m`

Explanation:

The given information means,he can throw the ball with maximum velocity of `u` at an angle of 45 degrees ,so that it completes a projectile motion which will have a range of `100m`

Range of a projectile motion is given as,`R= (u^2 sin 2theta)/g`

Here, `R=100` and `theta =45`

So, we get,`u^2= 1000`

Now,if he would have thrown the ball with this velocity upwards,if it would have reached a maximum distance of `s`,

Then using `v^2 = u^2 -2gs` we get,

here, `v=0,u^2 = 1000`

So, `s=50 m`

Derivation of formula of range of projectile motion.

Suppose,the projectile projected with a velocity of `u` at an angle of `theta` w.r.t horizontal, then if horizontally it reaches the maximum distance in time `T`, so,we can write, `R=u cos theta*T`..1

And,from vertical motion we can say,if it reaches the maximum height in time `t`,then,`t=(u sin theta)/g` (using `v=u-g t`,here, `v=0 .u=u sin theta`)

Now,the projectile will take same time to fall down, So, `T=2t=(2u sin theta)/g`

So,putting this value of `T` in 1 we get,

`R= (u cos theta) *(2u sin theta)/g = (u^2 sin 2 theta)/g`