Question

A ball is gently droped from a height of 90m. If the velocity increase uniformly at the rate of 10ms-2 with what velocity will it strike the ground?After what time will it striked the ground?​

Answer 1

Answer:

42.42m/s

Explanation:

u = 0m/s

h = 90 m

g = 10 ms⁻²

v² = u² + 2gh

v = √2x10x90

   =√1800

    = 42.42m/s

Answer 2

Answer:

42.42 m/s and 6.7 second

Explanation:

Given height= 90m.

g taken as = 10m/s²

u= 0 m/s

v= ?

t= ?

Using the equation 2as= v² − u²

2 × 10 × 90= v²

1800=v²

v = √1800

= 30√2

≈ 42.42 m/s

Now using the equation s= ut + 1/2at²

90= 0×t + 1/2 × 10 × t²

90= 5t²

t²=45

t= √45

≈ 6.7 sec