Question

A ball is thrown up with a velocity of 19.6m/s. How long will it take to reach the maximum height. How high will it go

Answer 1

g=-9.8
u=-19.6
v=0
t=v-u/a
t=-19.6/-9.8=2s
s=ut+1/2at²
=19.6*2+(-9.8*2)
9.8m
is this right

Answer 2

Maximum height of 19.6 m will be reached by the ball in 2 seconds.

•According to the equation of motion

v = u + at

when v will equal to zero, maximum height will attain at that time t.

•So, put v = 0 we have

t = -u/a

put u = 19.6 and a = -g , we have

t = 19.6/9.8

t = 2 seconds

•by using other equation of motion

as s = ut + 1/2 at²

we have

s = (19.6×2)+(1/2)(-9.8)2²

s = 19.6 m

•Hence, maximum height of 19.6 m will be reached by the ball in 2 seconds.