a ball thrown up vertically returns to the thrower after six seconds. find the velocity with which it was thrown up ,the maximum hight it reaches, and the position after four seconds
Answers
Answer:
Initial velocity is 29.4 m.
Explanation:
Given,
Total time taken( to reach the highest point & to return ) is 6 sec.
It's going the direction opposite to gravity, with no horizontal velocity. Thus, let the initial velocity ( vertical ) be u.
Velocity( final velocity ) at the highest point will be 0, and time taken to reach that point is half of the total time.
Using the equations of motion :
- v = u + at
= > 0 = u -gt { acceleration is against the gravitational acceleration , acceleration = - g }
= > u = gt
= > u = ( 9.8 m/s^2 )( 6/2 s )
= > u = ( 9.8 m/s^2 ) 3 s
= > u = 29.4 m/s
- v^2 = u^2 + 2aS
= > 0 = u^2 - 2gS
= > u^2 = 2gS
= > ( gt )^2 = 2gS
= > gt^2 = 2S
= > ( 9.8 )( 3 )^2 m = 2S
= > 44.1 m = S
When returning to the initial position : from the highest point, now, initial velocity is 0.
According to the total system, 4th second is the 1st second for the second one.
= > S = ut + 1 / 2 at^2 m
= > S = 0 + 1 / 2 (g 1^2 ) m
= > S = 1/2( 10 ) m
= > S = 5 m.
Answer:
Explanation:
The ball returns to the ground after 6 seconds.
Thus the time taken by the ball to reach to the maximum height (h) is 3 seconds i.e t=3 s
Let the velocity with which it is thrown up be u
(a). For upward motion,
v=u+at
∴ 0=u+(−10)×3
⟹u=30 m/s
(b). The maximum height reached by the ball
h=ut+ 1 /2 at2
h=30×3+ 1/2 (−10)×3 2
h=45 m
(c). After 3 second, it starts to fall down.
Let the distance by which it fall in 1 s be d
d=0+ 1/2 at 2 ′
where t =1 s
′
d= 1/2×10×(1) 2
=5 m
∴ Its height above the ground, h
′
=45−5=40 m
Hence after 4 s, the ball is at a height of 40 m above the ground.