A projectile is shot from the edge of a cliff h =205 m above ground level with a

Question

A projectile is shot from the edge of a cliff h =205 m above ground level with an initialspeed of v0 = 135 m/sat an angle of 37.0° with the horizontal. (a) Determine the time taken by the projectile to hit pointP at ground level.
(b) Determine the range X of the projectile as measuredfrom the base of the cliff. (c) At the instant just before the projectile hits pointP, find the horizontal and the vertical components of itsvelocity. (Take up and to the right as positive directions.) (d) What is the the magnitude of the velocity?
(e) What is the angle made by the velocity vector with thehorizontal? A projectile is shot from the edge of a cliff h =205 m above ground level with an initialspeed of v0 = 135 m/sat an angle of 37.0° with the horizontal. (a) Determine the time taken by the projectile to hit pointP at ground level.
(b) Determine the range X of the projectile as measuredfrom the base of the cliff. (c) At the instant just before the projectile hits pointP, find the horizontal and the vertical components of itsvelocity. (Take up and to the right as positive directions.) (d) What is the the magnitude of the velocity?
(e) What is the angle made by the velocity vector with thehorizontal?

Explanation / Answer

height h = 205 m Initial speed v = 135 m / s angle = 37 degrees In vertical direction : -------------------- Initial velocity u = v sin 37                         = 81.24 m / s Accleration a = -9.8 m / s ^ 2 displacement S = h = - 205 m from the relation S = ut + ( 1/ 2) at ^2                -205 = 81.24 t -4.9 t ^ 2       4.9t^2 -81.24 t - 205 =0 from this t = {81.24 ± [ (-81.24)^2-4(4.9)(-205) ] } / 2 ( 4.9)                = [ 81.24 ± 103.04] / 9.8                = 92.14 s    Since time is notnegatvie (b). Horizontal range R = t * v cos 37                                    = 9934.32 m (c). horizontal component   = v cos 37 = 107.81 m /s vertical component = v sin 37 + gt

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