Question

Pls solve for me Dont copy from internet I adk Brainiacs to help me​

Answer 1

Hey there!

Time taken to attain maximum height = 8 / 2 = 4 sec

Now,

a) when a ball is thrown vertically upward,

Final Velocity(v) at max. height = 0 m/s

g = - 9.8 m/s²

From,

First Equation of motion,

v = u + at

0 = u +(- 9.8 * 4)

0 = u - 39.2

u = 39.2 m/s

Hence, the initial velocity by which the ball was thrown is 39.2 m/s.

b) The maximum height it reaches would be:

From third equation of motion:

v² = u² + 2aS

0 = (39.2)² + (2* - 9.8*S)

0 = 39.2 * 39.2 - 2*9.8S

- (39.2 *39.2) / - 2 * 9.8 = S

or, S = 78.4 m

Hence, the maximum height is reaches it 78.4 m,

c) t = 5 sec

It reaches maximum height in 4 sec. and for 1 sec it falls from the top.

So, u = 0 m/s

a = 9.8 m/s²

From second equation of motion,

S = ut + 1/2 at²

S = 0 + 1/2 * 9.8 * (1)²

S =  4.9 m

So, after five second it will be at 4.9 below the maximum height.

Or, after 5 s, it will be at height = 78.4 - 4.9 = 69.5 m from  the ground.

Hope It Helps You!