a ball is thown up vertically returns to the thower after 6 secs. find the velocity with which it was thown up, the maximum height it reaches and its position after 4 secs.
Answers
at highest point velocity will be zero
Let time taken to reach highest point = t
use , kinematic equation ,
V = U + at.
0 = U - 9.8t
U = 9.8t
A/C to question ,
time taken to return the ball = 6
hence, time of ascent + time of descent = 6 sec
but we know ,
time of ascent = time of descent
so,
time of ascent ( t) = 6/2 = 3sec
hence,
initial velocity of ball = 9.8t = 29.4 m/s
maximum hight = u^2 /2g = (29.4)^2/19.6
= 44.1 m
position at 4sec :
S = Ut + 1/2at^2
= 29.4*4 - 1/2 *9.8*(4)^2
= 117.6 - 78.4
= 39.2 m
_/\_Hello mate__here is your answer--
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The ball takes a total of 6 s for its upward and downward journey. (Time of ascent is equal to the time of descent.)
Hence, it has taken 3 s to reach at the maximum height.
v = 0 m/s
g = −9.8 ms−2
Using equation of motion,
v = u + at
⇒0 = u + (−9.8 × 3)
⇒ u = 9.8 × 3 = 29.4 m/s
Hence, the ball was thrown upwards with a velocity of 29.4 m/s.
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Let the maximum height attained by the ball be h.
u = 29.4 m/s
v = 0 m/s
g = −9.8 ms−2 (upward direction)
Using the equation of motion,
s = ut +1/2 gt^2
⇒ hℎ = 29.4 × 3 −1/2 × 9.8 × 32
⇒ ℎh = 44.1 m
Hence, the maximum height is 44.1 m.
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Ball attains the maximum height after 3 s. After attaining this height, it will start falling downwards.
In this case,
u = 0 m/s
Position of the ball after 4 s − 3 s = 1 s can calculate by Using the equation of motion,
s= ut +1/2gt^2
⇒ h= 0 × 1 +1/2 × 9.8 × 12
⇒ h= 4.9 m
Now, total height = 44.1 m This means, the ball is 39.2 m (44.1 m − 4.9 m) above the ground after 4 seconds.
I hope, this will help you.☺
Thank you______❤
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