A projectile is shot from the edge of a cliff 185 m above ground level with an i

Question

A projectile is shot from the edge of a cliff 185 m above ground level with an initial speed of vo=125 m/s at an angle of 35.0° with the horizontal, as shown in the figure.

Find the following:

35.0 (a) Determine the time taken by the projectile to hit point P at ground level (b) Determine the distance X of of point P from the base of the vertical cliff. km (c) At the instant just before the projectile hits point P, find the horizontal and the vertical components of its velocity. (Assume the positive directions are upward and to the right.) m/s (horizontal) m/s (vertical) (d) At the instant just before the projectile hits point P, what is the magnitude of the velocity? m/s (e) At the instant just before the projectile hits point P, what is the angle made by the velocity vector with the horizontal? o below the horizontal (f) Find the maximum height above the cliff top reached by the projectile

Explanation / Answer

Here ,

angle , theta = 35 degree

vo = 125 m/s

height of cliff , h = 185 m

a) let the time taken to reach P is t

Using second equation of motion

h = u * t + 0.5 * a * t^2

-185 = 125 * sin(35) * t - 0.5 * 9.8 * t^2

solving for t

t = 16.9 s

the time taken to travel to P is 16.9 s
b)

distance , x = vo * cos(35) * t

distance , x = 125 * cos(35) * 16.9 s

x = 1730 m

the point P is 1730 m away

c)

horizontal velocity = vo * cos(35)

horizontal velocity = 125 * cos(35)

horizontal velocity = 102.4 m/s

vertical velocity = vo * sin(theta) - g * t

vertical velocity = 125 *sin(35) - 16.7 * 9.8

vertical velocity = -92 m/s

the vertical velocity of the object is - 92 m/s

d)

magnitude of velocity = sqrt(vx^2 + vy^2)

magnitude of velocity = sqrt(92^2 + 102.4^2)

magnitude of velocity = 137.6 m/s

the magnitude of velocity at point P is 137.6 m/s

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