Question

6. A ball is thrown vertically upwards with speed u m s-¹ from a point P at height h metres above the ground. The ball hits the ground 0.75 s later. The speed of the ball immediately before it hits the ground is 6.45 m s. The ball is modelled as a particle. (a) Show that u = 0.9 (3) (b) Find the height above P to which the ball rises before it starts to fall towards the ground again. (2) (c) Find the value of h. (3)​

Answer 1

a)u = 0.9 m/s

b)0.04m

c)2.08125 m

Explanation:

a) v = u + at

-6.45 = u + (-9.8)(0.75)

-6.45 = u - 7.35

u = 0.9 m/s

b) vt^2 - v0^2 = -2gh

here vt =0

hence , h= vt^2 - v0^2 / -2g

= 0 - (-0.9)^2 / -2 × 9.8

= -0.81 / -19.6

≈ 0.04m

c) h = {( v0+ vt) /2} × t

= {(-0.9 + 6.45) / 2 } × 0.75

=
`(5.55*0.75)/(2)`

= 2.08125 m