A ball is thrown vertically upwards from the ground level with a velocity of 19.6m/s. How long does it take to reach the maximum height? What maximum height does it reach? What is the velocity of the ball at t=3s (Please send me the solution with an attachment)
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The ball returns to the ground in 5 seconds. This means that the ball takes 2.5 seconds to attain the maximum height. Let me check if this information is correct. Applying v=u+at, (at maximum height, the speed of the ball, v=0 and a=−g=−9.8 ms−2) we can find that the time required to reach maximum height is 2 seconds. Hence, the given information is wrong. But, we can still solve it using another formula.
Now, applying the formula v2=u2+2as, (here, a=−g=−9.8 ms−2, s=h , maximum height) :
02=(19.6)2+2⋅(−9.8)⋅h⟹h=19.6
∴ Maximum height attained by the ball is 19.6 m.
Now, applying the formula v2=u2+2as, (here, a=−g=−9.8 ms−2, s=h , maximum height) :
02=(19.6)2+2⋅(−9.8)⋅h⟹h=19.6
∴ Maximum height attained by the ball is 19.6 m.
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hey mate here is your answer ....
v=u+at
[here u=19.6m/s,a=g=-9.8m/s,and at highest point v=0]
0=19.6-9.8×t
t=2 sec..
v^2-u^2=2ah
0-19.6×19.6=2×(-9.8)×h
h=19.6m
after 3 sec ball will come down words direction.
v=u+at
v=19.6-9.8×3
v=-9.8m/s
here negative sign suggest us that ball will come down word direction...
hope it's helpful to you....
v=u+at
[here u=19.6m/s,a=g=-9.8m/s,and at highest point v=0]
0=19.6-9.8×t
t=2 sec..
v^2-u^2=2ah
0-19.6×19.6=2×(-9.8)×h
h=19.6m
after 3 sec ball will come down words direction.
v=u+at
v=19.6-9.8×3
v=-9.8m/s
here negative sign suggest us that ball will come down word direction...
hope it's helpful to you....
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