A ball thrown vertically up return to the thrower after 6s find the volocity with which it was thrown up
Answers
Answer:
Time to reach Maximum height ,
t = 6/2 = 3 s.
v = 0 (at the maximum height )
a = - 9.8 m s-²
◆ a) Using, v = u + at, we get
0 = u - 9.8 × 3
or, u = 29.4 ms-¹
Thus, the velocity with which it was thrown up = 29.4ms-¹
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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.