(a) Determine the time taken by the projectile to hit point P at ground level. _
ID: 2021090 • Letter: #
Question
(a) Determine the time taken by the projectile to hit point P at ground level.
______________s
(b) Determine the range X of the projectile as measured from the base of the cliff.
______________km
(c) At the instant just before the projectile hits point P, find the horizontal and the vertical components of its velocity. (Take up and to the right as positive directions.)
______________horizontal m/s
______________vertical m/s
(d) What is the the magnitude of the velocity?
______________m/s
(e) What is the angle made by the velocity vector with the horizontal?
______________° (below the horizontal)
(f) Find the maximum height above the cliff top reached by the projectile.
_______________m
Explanation / Answer
(f) max height given by max. height=(Vo • sinT)² / (2g) = (155m/s•sin37º)² / (2•9.8m/s²) = 444 m (a) time to reach max height can be determined via s = 444m = ½at² = ½•9.8m/s²•t² = 4.9m/s²•t² t = 9.5s So to rise and then fall 444m takes 19 s. The initial vertical velocity = Vv = 155m/s • sin37º = 93.3 m/s This will be the downward velocity when the projectile returns to cliff height. The velocity at the bottom can be determined via Torricelli's: v² = u² + 2as = (93.3m/s)² + 2•9.8m/s²•205m = 12719 m²/s² v = 112.8 m/s Then this portion of the flight time is v = 112.8 m/s = u + at = 93.3m/s + 9.8m/s²•t t = 2s So the total flight time is 19s + 2s = 21 s (b) horizontal velocity Vh = V•cosT = 155m/s•cos37º = 123.8 m/s Then the range x = Vh • t = 123.8 m/s • 21s = 2600m = 2.6 km (c) already done: Vh = 123.8 m/s; Vv = -112.8 m/s (d) V = 167.5 m/s (e) T = arctan(-112.8 / 123.8) = -42.3º (42.3º below the horizontal)
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