From a point on the ground ,at a distance 15m,from the foot of a vertical wall,a ball is thrown at an angle of 45 degree ,which just clears the top of the wall afterwards strikes the ground at a distance 5m on the other side.Find the height of the wall.
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82
Total range = 15 + 5 = 20
Horizontal velocity = range/ time of take off
U cos45 degree = 20 x g / 2 U sin45 degree
U= 14 m/s
Time= distance/ velocity
15/ 14 cos 45 degree
= 0.50 sec
Displacement= h= U sintheta x t - 1/2 gt²
h = 14 sin 45*0.50 - 4.9*0.5^2
h = 3.72 m/s
Horizontal velocity = range/ time of take off
U cos45 degree = 20 x g / 2 U sin45 degree
U= 14 m/s
Time= distance/ velocity
15/ 14 cos 45 degree
= 0.50 sec
Displacement= h= U sintheta x t - 1/2 gt²
h = 14 sin 45*0.50 - 4.9*0.5^2
h = 3.72 m/s
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8
use equation of trajectory to do this one and it will be easier
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