A ball is thrown upwards takes 4 seconds to reach maximum height.
Find-1 _initial speed with which the ball
was thrown.
2_maximum height reached
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given,
time taken = 4second
gravity = -9.8m/s²
final velocity(v)=0
initial velocity (u) =?
we know that,
v=u+gt
0=u+(-9.8)*3
-u= -29.4
so, u=29.4m/s
now,
let h be the height of ball
then,
h=ut+1/2at²
= 29.4*4+ 1/2*(-9.8)*4²
= 117.6+1/2*(-9.8)*16
= 117.6+8*(-9.8)
=117.6-78.4
=39.2m
Thus,h=39.2m
time taken = 4second
gravity = -9.8m/s²
final velocity(v)=0
initial velocity (u) =?
we know that,
v=u+gt
0=u+(-9.8)*3
-u= -29.4
so, u=29.4m/s
now,
let h be the height of ball
then,
h=ut+1/2at²
= 29.4*4+ 1/2*(-9.8)*4²
= 117.6+1/2*(-9.8)*16
= 117.6+8*(-9.8)
=117.6-78.4
=39.2m
Thus,h=39.2m
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