Question

A ball is projected from ground.after 2 s ,the velocity of ball makes 45°angle with horizontal and after next 2 s ,it moves horizontal.the velocity of projection of ball is :

Answer 1

For answer see below image

If correct then please marks as brainliest

It is m/s instead of m sorry for mistake

Answer 2

Answer:

velocity of ball is 20 sqare root 5meters per sec

Explanation:

After 4 sec ball is at highest Point ,

that is the total time taken to reach

$\displaystyle \frac{u sin\theta}{g} = 4$

$[u sin\theta = 40 - - - - - 1$

the velocity of ball makes 45°angle with horizontal

$\displaystyle tan45^o  = \frac{u sin\theta – g t}{u cos\theta}$

Here g gravity=10mts per sec

$tan45^o  = \frac{40 – (10 \times 2)}{u cos\theta}$

$\displaystyle u cos\theta = 20 - - - - 2$

squaring and adding 1 and 2

we get

$u^2 (sin^2 \theta + cos^2 \theta) =$

$40^2 + 20^2$

$u^2  = 2000$

$u = 20 \sqrt{5} m/s$

velocity of projection of ball is :

$u = 20 \sqrt{5} m/s$

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