Please help... Ive spent so long trying to figure this out.

Question

The question is...

 

A projectile is shot from the edge of a cliff h = 285 m above ground level with an initial speed of v0 = 125 m/s at an angle of 37.0° with the horizontal, as shown in the figure below

 

(a) Determine the time taken by the projectile to hit point P at ground level.

(b) Determine the range X of the projectile as measured from the base of the cliff.

(c) At the instant just before the projectile hits point P, find the horizontal and the vertical components of its velocity. (Take up and to the right as positive directions.)
horizontal m/s
vertical m/s

(d) What is the the magnitude of the velocity?

(e) What is the angle made by the velocity vector with the horizontal?

(f) Find the maximum height above the cliff top reached by the projectile.

 

Answer 1

Hey Not.

 

a)

The angle of 37 deg will effect your initial horizontal velocity and initial vertical velocity. A 37 degree angle is really convenient because sin(37) = 3/5, and cos(37) = 4/5.

 

Your v0 at the angle given is 125m/s. If you break that into horizontal (think cosine) and vertical (think sine) components:

 

v0x = 125*4/5 = 100 m/s

v0y = 125*3/5 = 75 m/s

 

To figure out the time it takes to reach the ground, you need to refer to your projectile motion equation:

 

y = -1/2*g*t^2 + v0y*t + y0

 

Since your units are reported in m/s, g = 9.8 m/s^2

We determined v0y above to be 75 m/s

y0 will be the height of your cliff: 285 m

 

So

 

y = -1/2*9.8*t^2 + 75*t + 285

 

y = -4.9t^2 + 75t + 285

 

Since you want to know when the ball hits the ground, that means y = 0.

 

0 = -4.9t^2 + 75t + 285.

 

This is a simple enough quadratic to solve using the quadratic formula, or if you're lazy, Wolfram Alpha.

 

t = -3.15 sec, 18.46 sec

 

Time obviously can't be negative, so we can ignore that. Your projectile was in the air for 18.46 seconds.

 

b)

The range of the projectile is simple to calculate once you know how long its been in the air.
The horizontal component of the velocity will not be effected by outside forces, since physicists love to negate air resistance in their calculations.

 

distance = rate*time

R = v0x*t

R = 100m/s * 18.46 sec  = 1,846 meters.

 

c)

Horizontal component doesn't change throughout the projectile's motion: 100m/s

 

Vertical component is a little more tricky. If the projectile is launched up with a vertical velocity of 75 m/s, it will be traveling 75 m/s when it comes back down and reaches its starting point. Your projectile wasn't fired from the ground, however; it was fired from 285m up. So it still has some ways to go, and gravity will be accelerating it all the way down. You have one issue, though: you don't know how long it took for the projectile to fall back down to its original starting height yet. All you know is how long it took to for it to reach the ground, which was 18.46 seconds.

 

To find how long it took to reach its original starting height, use the same formula as above:

 

y = -1/2*g*t^2 + v0y*t + y0

 

Since you're starting at 285m and want to end at 285m, y and y0 will both be 285m, so they will cancel out to zero.

 

0 = -4.9t^2 + 75t

0 = t(-4.9t + 75)

t = 0, 15.31

 

We aren't concerned with the 0, since that's obviously the starting point. It took the projectile 15.31 seconds to go from its starting height all the way to its peak and back down to its starting height.

 

Since the entire trajectory required 18.46 seconds, the last bit of its free-fall was 18.46 - 15.31 = 3.35 seconds long. This is how long gravity was able to act on it. Additionally, since your object is now initially moving downward, its velocity will be -75.

 

velocity = -gt + v0y

 

V = -9.8*3.35 + (-75)

V = 107.83 m/s

 

d)

Magnitude is found by doing vector sums, which looks very similar to the Pythagorean Theorem.

 

M^2 = 107.83^2 + 100^2

 

M = 147.06

 

e)

Angle with the horizontal can be found by taking arccos(horizontal/magnitude)

 

arccos(100/147.06) = 0.823 rad, or 47.16 degrees. Looks like its fall got a little steeper since its y-component increased a bit, which we expect.

 

f)

Maximums can be found by figuring out what -b/2a is in your quadratic then plugging it back into your equation.

 

y = -4.9t^2 + 75t + 285

 

a = -4.9

b = 75

c = 285

 

tmax = -b/2a = -75/9.8 = 7.65 sec (this is half of the 15.31 secs we found earlier)

 

ymax = -4.9*(7.65)^2 + 75*(7.65) + 285

 

ymax = 572 m

 

Hope this helps.