Question

A mortar shell is launched with a velocity of 100 m/s at an angle of 30 degrees above horizontal from a point on a cliff 50 meters above a level plain below. How far from the base of the cliff does the mortar shell strike the ground?

Answer 1

First, calculate the x and y components of initial velocity:

v0x = 100 cos 30 = 86.6 m/s
v0y = 100 sin 30 = 50 m/s

Then, solve for total time in air:

y = y0 + voy(t) - (0.5)gt^2
0 = 50 + (50 m/s)t - (0.5)(9.8 m/s^2)t^2
Quadratic Equation:
(-50 +/- sqrt(50^2 - (4*-4.9*50))) / (2*-4.9)
t = 11.12 s

Now, we can solve for range:

x = (v0x)(t)
x = (86.6 m/s)(11.12 s)
x = 963 m

Answer 2

Thank you

Answer 3

yw ;)