# A ball is thrown and projected upward with angle of elevation with respect to the horizontal. The ball reaches its maximum height in 1.5 seconds and covers a horizontal distance of 200 ft....

- A ball is thrown and projected upward with angle of elevation with respect to the horizontal. The ball reaches its maximum height in 1.5 seconds and covers a horizontal distance of 200 ft. Calculate the initial velocity of the and the angle of elevation it makes with the horizontal.

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### 1 Answer

Let the angle of projection with the horizontal be `theta` .

Then the vertical component of velocity is `usintheta` , and the horizontal component of velocity is `ucostheta` , where u is the initial velocity of projection.

It reaches a maximum height in 1.5 seconds.

Analyzing the vertical component, at the maximum height

`0=usintheta-g*1.5`

`rArr usintheta=1.5*32.2` --- (i)

Again considering the horizontal component, horizontal distance travelled by this time is `ucostheta*1.5` .

This distance is half its horizontal range.

So, `ucostheta*1.5=1/2*200`

`rArr ucostheta=100/1.5=200/3` --- (ii)

(i)/(ii), `tantheta=(1.5*32.2)/(200/3)=0.7245`

`rArr theta=arctan(0.7245)=35.92^o~~36^o`

And `costheta=1/(sqrt(1+tan^2theta))=1/(sqrt(1+0.7245^2))`

`=0.8098`

Hence, `u=200/3/costheta=200/(3*0.8098)`

`=82.3` ft/sec.

Therefore, the initial velocity of projection was 82.3 ft/sec. and angle of projection was `36^o` .

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