Question

Please answer this question urgent. ....

Answer 1

For 2nd question

The ball is thrown vertically upward so x component of velocity is 0.

At the height of 45m velocity of projection is 0 so

V = 0

U = ?

S = 45

a = -g

v^2 = u^2 + 2aS

0 = u^2 +2*(-10)*45

900 = u^2

u = 30m/s upward

Now we have to check that when the velocity will be 15m/s. The velocity will be 15m/s two time i.e. when the ball is going upward and downward, so we will check the position of the ball at which its speed is half

v^2 = u^2 + 2aS

15*15=30*30 + 2* -10 *S

20S=900–225

20S=675

S = 675/20=33.75 m from ground so

S=ut+1/2at^2

33.75=30*t-5t^2

By solving this quadratic we will get

t= 4.5s and t =1.5s

At this both instant the velocity will be half of the initial velocity of the projection .

Hope it helps u.