Question

A ball is thrown upwards with a speed of 39.2 m/s. Calculate (a) the maximum height it reaches, and (b) the time taken in reaching the maximum height.

Answer 1

max height reached=u^2÷2g=39.2x39.2÷2x9.8=78.4
time taken to reach max height=u÷g=39.2÷9.8=4

Answer 2

Answer:

Explanation:

intial velocity(u) = 39.2 m/s

final velocity(v) = 0

acceleration(a) = -9.8 m/s2

a) using third equation of motion

2as = v^2 -u^2

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

-19.6 s = - 1536.64

19.6s = 1536.64

s = 1536.64 / 19.6

= 78.4 m

so,

the maximum height covered = 78.4 m

b) let time taken be t

using acceleration formula

a= (v-u)/t

-9.8 = (0- 39.2)/t

-9.8 t = -39.2

9.8t = 39.2

t = 39.2 / 9.8

= 4 s

so,

the time taken was 4s to reach maximum height.