Question

A projectile is shot from the edge of a cliff a 125m above ground level with an initial speed of 65m/s at an angle of 37^o below the horizontal.

Find the y-component of the VELOCITY just before it reaches the bottom of the cliff

Find the magnitude and angle of the displacement for the entire trip from cliff to the ground

Answer 1

given
h = 125 m
vo = 65 m/s
theta = 37 degrees

initial components,
voy = -vo*sin(37)
= -65*sin(37)
= -39.1 m/s

vox = vo*cos(37)
= 65*cos(37)
= 51.9 m/s

let vy is the y component of the velocity

vy^2 - voy^2 = 2*g*h

vy = sqrt(voy^2 + 2*g*h)

= sqrt(39.1^2 + 2*9.8*125)

= -63.1 m/s <<<<<<<<<<------------------Answer

let t is the time taken to reach the grund.

use,

h = voy*t + (1/2)*g*t^2

125 = 39.1*t + (1/2)*9.8*t^2

on solving the above equation

we get, t = 2.447 s

displacement in x-direction,

x = vox*t

= 51.9*2.447

= 127 m

magnitude of total displacemet = sqrt(h^2 + x^2)

= sqrt(125^2 + 127^2)

= 178 m <<<<<<<<<<----------------Answer

theta = tan^-1(h/x)

= tan^-1(125/127)

= 44.5 degrees <<<<<<<<<<----------------Answer