Question

A ball is thrown upwards takes 4 seconds to reach the maximum height find the initial speed with which it was thrown and the maximum height reached

Answer 1

Answer:

time taken (t)= 4s

final velocity (v)=0

acceleration (a) = -9.8 ( as it is thrown upward)

a)

let inital velocity be u

using first equation of motion

v = u + at

0 = u + (-9.8)(4)

0 = u - 39.2

39.2 = u

so,

the initial velocity was 39.2 m/s

b)

let maximum height be s

using third equation of motion

2as = v^2 -u^2

2 (-9.8)(s) = 0^2 - (39.2)^2

-19.6 s = - 1536.64

19.6 s = 1536.64

s = 1536.64/19.6

78.4

so, the height covered was 78.4 m

Explanation:

Answer 2

Answer:

Explanation:

T= u/g...u= 20m/s

H = u^2/2g= 400/20=20m