Question

a body is thrown up with a velocity 50 m/s. calculate the maximum height reached and the time taken to reach thebmaximum heght.(a = - 10 m/s)

Answer 1

Explanation:

initial velocity, u = 50m/s

final velocity, v = 0m/s

acc due to gravity, g = -10m/s^2

let max height be h

let time taken be t

Using 3rd equation of motion

${v}^{2} - {u}^{2} = 2gh$

0-2500 = -20h

h = 125m

using 1st equation of motion

v = u + gt

0 = 50 - 10t

t = 5s

Answer 2

Answer:

initial velocity = 50m/s(u)

final velocity = 0m/s(v)

a= -10m/s

time = ?

distance = ?

time = v-u=at

0-50 = -10t

-50/-10=5sec =time

s= ut+1/2*at^2

s= 50*5+1/2*-10*25

s=250-125

s=125m