Question

A ball thrown upwards takes 4 s to reach the maximum height. Find (a) the initial speed with which it was thrown, and (b) the maximum height 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

Answer 2

time taken = 4 sec

final velocity = 0

acceleration = -9.8

a) v = u + at

0 = u+(-9.8)(4)

0 = u - 39.2

39.2 m/s = initial velocity

b) maximum height = s

2as = v^2 - u^2

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

19.6 × s = 1536.64

s = 1536.64/19.6

s = 78.4 m

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