Question

a ball is projected obtiquely in air with 100m/sec. ratio of time taken for travelling max horizontal distance to time taken for travelling half of the max horizaontal distance

Answer 1

We know that R = u2 * Sin 2θ / g.

As, the range is maximum, so, θ = 450

So, 80 = u2 /g.

Now, H = height reached by ball = u2 Sin2θ / 2g.

Putting value of θ = 450 , u2 / g = 80, we get

H = 80/4 = 20m

Please approve it if you like......

Answer 2

max. horizontal distance is (u^2 sin 2thita)/g

total tym taken = (2u sin thita )/ g
tym taken for half max distance =( u sin thita )/g

so ratio is 2:1