Question

A projectile is fired with an initial speed of 200 m/s, angle of elevation 60 degrees, and fired from 100m above the ground.

What is the range, maximum height, and speed at impact?

Answer 1

Fundamental kinematics equation ought to take care of this one:
\[\Delta s=vt+0.5at^2\]

Answer 2

I recommend treating the horizontal and vertical directions separately by taking vector components as well.

Answer 3

We haven't really talked much about physics in this class (multivariable) and I haven't taken basic physics for awhile. What would be acceleration here?

Answer 4

at maximum height velocity v = 0
vertical component of velocity at start is 200*sin 60 m/s
use
v^2 = u^2 + 2as where s = max height, and a = -9.81m s^-2

so

0 = (200*sin60) ^2 - 2*9.81* s
solve this for s

Answer 5

If we're assuming that the projectile is pretty near to Earth's surface, you can use -9.81 m/s/s as gravitational field strength (acceleration).

Answer 6

acceleration = gravity = - 9.81 m s_2

Answer 7

Also, always draw a picture!