Question

A ball is projected vertically up with a speed of 50m/s. Find the maximum height the time to reach the maximum height, and the speed at the maximum height?

Answer 1

Solution :

Initial speed= u=50m/s

g=10m/s2

At maximum height V=0 m/s

Maximum height:

V²-u²=2as

0²-50²=2*-10*s

s=50*50/20

s=125m

∴maximum height reahed =125m

b) time taken to reach maximum height

t=u/g

t=50/10=5 sec

∴Time taken to reach maximum height =5 sec

c) the speed at half the maximum height:

s=125/2=62.5m

v=?

u=50m/s

a=-10m/s²

From  v²-u²=2as

V=√u²+2as

V=√50²+2x(-10)x 62.5

v=√2500-1250

v=√1250=35 m/s

Answer 2

Hi,

Thank you for posting your question on embibe. Hope your prepration is going well.

Here is the answer to your question

U = 50 m/s, g = –10 m/s2 when moving upward, v = 0 (at highest point). a) S = (V^2-u^2)/2a = (0-〖50〗^2)/(2(-10)) = 125 m maximum height reached = 125 m b) t = (v – u)/a = (0 – 50)/–10 = 5 sec c) s’ = 125/2 = 62.5 m, u = 50 m/s, a = –10 m/s2, v2 – u2 = 2as ⇒ v = √(( u^2+2as)) = √(50^2+2(-10)(62.5) ) = 35 m/s.