A projectile is shot from edge of cliff 140m above ground level with initial velocity of 100m/s at angle of 37

Question

determine
a) time taken by projectile to hit a point C at ground level
b) the range X of projectile as measured from the base of the cliff
c) At the instant just before the projectile hits the point C, find the horizontal and vertical components of its velocity
d) Find its final velocity ( magnitude and direction)

Answer 1

again, break the velocity into x and y components

Vx =cos(37)100 = 79.86 m/s
Vy =sin(37)100 = 60.18m/s

a) Vf = at + Vi

we have a = -9.8m/s^2 and initial velocity = 60.18m/s but we don't have final velocity to find the time. So what we are going to do is to find it

Vf^2 = 2ad + Vi^2

Vf^2 = 2(-9.8)(0 - 140) + (60.18)^2
Vf = 79.78 m/s

now that we have final velocity, plug it in for Vf in the equation Vf = at + Vi. But remember, velocity is a vector and vector has directions. The final velocity when it hit the ground is downward, so it should be -79.78m/s

-79.78 = (-9.8)t + 60.18
t = 14.28s


b) X = vt
x = 79.86 * 14.28
x = 1140.4m


c) again, the vertical velocity and the horizonal velocity are independent of each other. The projectile keeps traveling at a constant horizonal speed the entire trip. so its horizonal speed is 60.18m/s and i just found its speed when it hits the ground in part a which is 79.78m/s


d) use pythegarian theorm

a^2 + b^2 = c^2
60.18^2 + 79.78^2 = c^2
c = 99.9 m/s^2

for the direction, because there is no specific detail that shows the projectile's direction when it's initially launched.
But assume it's lauched at 37 degrees North of East.

So the direction when the projectile hit the ground is South of East.

to find the angle use trig
theta = tan^-1 (79.78 / 60.18) = 52.97 degress

so the velocity is: 99.9 m/s at 52.97 degrees South of East

Answer 2

a 13.77 sec
b 1099.77 m
c.Vh= 79.863 m/s Vv= -300.31 m/s
d. Vfinal= 311.28 m/s