Question

a ball thrown upwards reach the maximum height. find a) the initial speed with which it was thrown and b) the maximum height it reached​

Answer 1

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