a)You throw a stone horizontally at a speed of 5.0 m/s from the top of a cliff that is 78.4 m high.
from above statement we got
height = 78.4 m
since the ball is thrown, so its vertical velocity would be zero
u = 0
taking g = 9.8m/s^2
now, using the equation of motion
h = ut + gt^2/2
now putting all the values in it
we got ,
78.4 = 9.8 * t^2/ 2
by solving we got,
t = 4 sec
b) now, since along the horizontal , no force acting and accelaration is zero so
R = ut , R is RANGE
R = 5 * 4
range = 20 m
c) vertical components of the stone’s velocity just before it hits the ground = v sin θ =
horizontal components of the stone’s velocity just before it hits the ground = v cos θ
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