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

Question

A projectile is shot from the edge of a cliff h = 125 m above ground level with an initial speed of v0 = 105 m/s at an angle of 37.0° with the horizontal, as shown in the figure below.

(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.)

(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

vo = 105 m/s

theta = 37

(a)

along vertical

initial velocity voy = vo*sin37

acceleration = a = -g

displacement y = -h

from equations of motion

y = voy*T + 0.5*gT^2

-125 = (105*sin37*T)-(0.5*9.8*T^2)

T = 14.6 s <<--------answer

+++++++++++++

(b)

X = vox*T = vo*cos37*T = 105*cos37*14.6 = 1224.3 m

++++++++++++

(c)

vx (horizantal) = vox+ax*T = (105*cos37)+0 = 83.8 m/s

vy (vertical) = voy + ay*T = (105*sin37)-(9.8*14.6) = -79.8 m/s

d)

magnitude = sqrt(vx^2+vy^2) = sqrt(83.8^2+79.8^2) = 115.7 m/s

(e)

angle = tan^-1(Vy/vx) = 43.6 ° (below the horizontal) <<----answer

(f)

at the maximum height vertical component becomes zero

vy = 0

from equations of motion

vy^2 - voy^2 = 2*ay*ymax

o-(105*sin37)^2 = -(2*9.8*ymax)

ymax = 203.7 m <<----answer

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