A projectile is shot from the edge of a cliff 115 m above ground level with an i

Question

A projectile is shot from the edge of a cliff 115 m above ground level with an initial speed of v0 = 47 m/s at an angle of 35.0 with the horizontal, as shown in the figure (Figure 1)Determine the time taken by the projectile to hit point P at ground level. Express your answer to three significant figures and include the appropriate units. t = SubmitMy AnswersGive Up Part B Determine the distance X of point P from the base of the vertical cliff. Express your answer to three significant figures and include the appropriate units. X = SubmitMy AnswersGive Up Part C At the instant just before the projectile hits point P, find the horizontal and the vertical components of its velocity. Express your answers using three significant figures separated by a comma. vx, vy = m/s SubmitMy AnswersGive Up Part D At the instant just before the projectile hits point P, find the magnitude of the velocity. Express your answer to three significant figures and include the appropriate units. v = SubmitMy AnswersGive Up Part E At the instant just before the projectile hits point P, find the angle made by the velocity vector with the horizontal. Express your answer to three significant figures and include the appropriate units. = below the horizon SubmitMy AnswersGive Up Part F Find the maximum height above the cliff top reached by the projectile. Express your answer to three significant figures and include the appropriate units. .

Explanation / Answer

We will neglect air resistance.
The initial vertical velocity is 47sin35 = 26.96 m/s
as velocity is acceleration by time and the vertical acceleration is g = 9.81 m/s/s
the time to peak point is found by
V = at
26.96 = 9.81t
t = 26.96 / 9.81 = 2.75 seconds

the height reached above the launch point is at 2.75 seconds
h = 1/2at^2
h = 1/2 (9.81)(2.75^2)
ANS f) h = 37.04 m

so the total drop from apex is 115 + 37.04 = 152.04 m
again using h = 1/2at^2 only this time to solve for t
152.04 = 1/2(9.81)t^2
t^2 = 37.89
t = 5.567 seconds
ANS a) total time in flight 5.567 + 2.75 = 8.317 seconds

the horizontal initial velocity never changes from 47cos35 = 38.5 m/s

ANS b) the horizontal distance is 38.5(8.317) = 320.2 m

the total straight line distance from the cliff top launch point is (320.2^2 + 115^2)^0.5 = 340.23 m


the vertical velocity component at landing will be
9.81(5.567) = 54.61 m/s
ANS c)
Vh = 38.5 m/s
Vv = 54.61 m/s

ANS d) total velocity is (38.5^2 + 54.61^2)^0.5 = 66.82 m/s

ANS e) angle is inv tan (54.61/38.5) = 54.82 degrees below horizontal

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