A ball was projected vertically from the top of a cliff of height 20m with...

A ball was projected vertically from the top of a cliff of height 20m with a velocity of 20m/s and was left to fall back to the base of the cliff, find the total time it spent in the air.?

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Aliyu
1 month ago
height of cliff(h)= 20m, initial velocity(u) = 20m/s g=10m/s a=90degrees H=maximum height
Let angle be a
Total height(S)= H+h
H=(U^2sin^2a)/2g
H=(20^2(sin^2 90))/(2*10)
H=(400 * 1^2)/20
H=20m
S=20+20=40m
Time taken to travel from top of cliff to H(t1) =(U sin a)/g
=(20sin90)/10
=20/10=2secs
To find time taken to travel from H to base of cliff, we substitute total height, 40m into h=ut+ 0.5gt^2
Initial velocity(u) is zero
40= 0(t) + 0.5(10)(t^2)
40= 5t^2
8= t^2
t= square root of 8
t=2.83secs
Total time taken to travel from top to base of cliff= 2+2.83= 4.83 secs
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agbojays
1 month ago
height of cliff(h)= 20m, initial velocity(u) = 20m/s g=10m/s a=90degrees H=maximum height
Let angle be a
Total height(S)= H+h
H=(U^2sin^2a)/2g
H=(20^2(sin^2 90))/(2*10)
H=(400 * 1^2)/20
H=20m
S=20+20=40m
Time taken to travel from top of cliff to H(t1) =(U sin a)/g
=(20sin90)/10
=20/10=2secs
To find time taken to travel from H to base of cliff, we substitute total height, 40m into h=ut+ 0.5gt^2
Initial velocity(u) is zero
40= 0(t) + 0.5(10)(t^2)
40= 5t^2
8= t^2
t= square root of 8
t=2.83secs
Total time taken to travel from top to base of cliff= 2+2.83= 4.83 secs
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isaaq
1 month ago