Q2. An arrow is shot from a height of 1.5 m toward a cliff of height H. It is shot with a velocity of 30 m/s at an angle of 65 degrees above the horizontal. It lands on the top edge of the cliff 4.0 s later. (a) What is the height of the cliff? (b) What is the maximum height reached by the arrow along its trajectory? (c) What is the arrow’s impact speed just before hitting the cliff? (d) If the same arrow was show at an angle of 35 degrees above the horizontal, where would the arrow land?
Please give step by step solution or write extremely neat. Thank you.
a)
using 2md equation of motion along vertical
h = 1.5 + 30* sin 65 * 4 - 4.9* 4^2
h = 31.857 m
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b)
max height
H = ho + ( u sin x) ^2/ 2g
H = 1.5 + (30 sin 65)^2 / 19.6
H = 39.217 m
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c)
v^2 = u^2 - 2 gh
v^2 = 30^2 - 2* 9.8* (31.857 - 1.5)
v = 17.464 m/s
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d)
horizontal distance of cliff
R = 30 cos 65 * 5 = 50.71 m
now if the angle is 35
then
time taken by the arrow to cover horizontal distance
t = 50.71 / 30 cos 35 = 2.06 s
vertical position of arrow
h' = 1.5 + ( 30 sin 35)^2/ 19.6
h' = 16.607 m
so the position of arrow wrt to ground is
(x, y) = ( 50.71, 16.607) m
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comment in case any doubt, will reply for sure.. goodluck
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