Question

A ball is thrown so that it just clears a 8 m wall 32 m away. If it left the hand 1.5m
above the ground and at an angle of 60° to horizontal, what was the initial velocity of
the ball?​

Answer 1

Answer:

Resolving into components we will find that horizontal velocity is v*cos600 whereas vertical component is v*sin600....(we are assuming the velocity of the ball as v initially)

So...forming equations let the ball reach the wall in time t.

so along horizontal

v*cos600*t=18 as distance between thrower and wall is 18 metres.

Along vertical,

using formula v2=u2-2*a*s..

here u being v*sin600 and v being 0,,a being g,i.e., 9.8m/s

s = 1.5m

we get another equation in v...

but...by only solving second equation....we get our required velocity as 39.2 m/s

Explanation:

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