Question

A ball is thrown vertically upward with a velocity of 5 m/s. It uniformly accelerates at the rate of -10m/s^2 , stops after reaching a certain height and begins to fall back downwards. Calculate the height attained by the ball before it falls back.

Answer 1

Howdy!!

your answer is ----

Given,
initial velocity u = 5m/s

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

and final velocity v = 0.

so, by the formula

v^2 = u^2 + 2gs

=> 0 = 5^2 -20s

=> -25 = -20s

=> s = 25/20= 1.25 meter

hence, the height attained by the ball before it falls back is 1.25 meter


hope it help you

Answer 2

hey buddy here is yur ans

we have
u= 5m/s
a= -10m/s^2
s=?
v=0
v= u+at
0= 5+(-10)*t
-5=-10*t
t= 1/2
s= ut+1/2at^2
s= 5*1/2+1/2(-10)(1/2)^2
s=2.5+(-5)(1/4)
s =2.5-1.25
s= 1.25 m
hope it help u