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 v_0 = 62 m/s at an angle of 35.0 degree with the horizontal, as shown in the figure (Figure1). 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. 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. At the just before the projectile hits point P, find the horizontal and the vertical components Express your answers using three significant figures separated by a comma.

Explanation / Answer

The initial vertical velocity is 62 sin(35) = 35.56m/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
35.56 m/s = 9.81t
t = 35.56 / 9.81 = 3.625seconds

the height reached above the launch point is at 3.625 seconds
h = 1/2at^2
h = 1/2 (9.81)(3.625^2)

h = 64.45 m

so the total drop from apex is 115 + 64.45m = 179.45 m
again using h = 1/2at^2 only this time to solve for t
179.45 = 1/2(9.81)t^2
t^2 = 36.62
t = 6.051 seconds
ANS a) total time in flight 6.051+ 3.625 = 9.676 seconds

the horizontal initial velocity never changes from 65cos35 = 50.78 m/s

ANS b) the horizontal distance is 50.78(9.676seconds) = 491.4m

the vertical velocity component at landing will be Vv= 62 sin(35)- gt

Vv= 35.56 - 9.81(9.676)

Vv= - 59.36m/s
ANS c)
Vh = 53.24 m/s
Vv =- 59.36 m/s

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