A projectile is shot from the edge of a cliff h=125m above ground level with an

Question

A projectile is shot from the edge of a cliff h=125m above ground level with an initial speed of Vo=145m/s at an angle of 37 degrees with the horizontal as shown. a)determine the time taken by the projectile to hit point P at ground level. b)determine the range X of the projectile as measured from the base of the cliff. Answer should be given in kilometers c)at the instant just before the projectile hits point P, find the horizontal and vertical components of its velocity (take up and to the right as positive directions). d)what is the magnitude of the velocity? e)what is the angle made by the velocity vector with its horizontal? f)find the maximum height above the cliff top reached by the projectile. Please bold the answer of each step

Explanation / Answer

Here ,

a)

let the time taken by the projectile is t

using second equation of motion

s = ut + 0.5 at^2

-125 = 145 * sin(37) * t - 0.5 * 9.8 * t^2

solving for t

t = 19.41 s

the time taken by the projectile to reach ground is 19.41 s

b)

Range = Vx * t

Range = 145 * cos(37) * 19.41

Range = 2247 m = 2.247 km

the range of the projectile is 2.247 km

c)

vertical component of velocity = 145 * sin(37) - 9.8 * 19.41

vertical component of velocity = -103 m/s

the vertical component of velocity is -103 m/s

horizontal component of velocity = 145 * cos(37)

horizontal component of velocity = 115.8 m/s

the horizontal component of velocity is 115.8 m/s

d)
magnitude = sqrt(vx^2 + vy^2)

magnitude = sqrt(103^2 + 115.8^2)

magnitude =155 m/s

the magnitude of the velocity is 155 m/s

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