physics help?

Question

A projectile is shot from the edge of a cliff h = 205 m above ground level with an initial speed of v0 = 115 m/s at an angle of 37.0° with the horizontal

(a) Determine the time taken by the projectile to hit point P at ground level.
s
(b) Determine the range X of the projectile as measured from the base of the cliff.
km
(c) At the instant just before the projectile hits the ground, 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?
m/s
(e) What is the angle made by the velocity vector with the horizontal?
° (below the horizontal)
(f) Find the maximum height above the cliff top reached by the projectile.
m

Answer 1

The key equations for motion under constant acceleration are:

Vf = Vi + A x T
D = Vi x T + (1/2) x A x T^2

where:
Vf = final velocity
Vi = initial velocity
A = acceleration
T = time interval
D = distance traveled

When the projectile is fired with velocity V at an angle X, the horizontal component of the velocity is V x cos X and the vertical component of the velocity is V x sin X.

Since we know V0 and the angle X we can determine the initial horizontal and vertical components of V0.

Since we are assuming that there is no friction with the air, and the only force on the projectile is gravity, which is vertical, the horizontal component of the projectile's velocity does not change throughout the problem.

The acceleration due to gravity is about 9.8 m/s^2.

At first, the vertical velocity of the projectile is up and gravity is slowing it down. Using the velocity equation with the known initial vertical velocity, the known acceleration due to gravity, and the known final vertical velocity, we can compute the time when the vertical velocity becomes 0. This is the time up.

Knowing this time, we can use the distance equation to compute the height above the cliff at the point where the vertical velocity is 0. This is the maximum height because after this the projectile starts falling.

We know the height of the cliff and the height of the projectile above cliff so we know the height above ground level. We also know the initial vertical velocity (0) and the vertical acceleration (gravity) so we can use the distance equation to find the time it takes for the projectile to go from its highest point to ground level. This is the time down.



Now we can start answering the questions:

a) We know the time up and the time down so we can add to get the total time.

b) The horizontal velocity is constant, so we can compute the horizontal distance traveled using the total time and the initial horizontal velocity.

c) We have the horizontal velocity component because it hasn't changed. Given that time down, we can use the velocity equation to compute the vertical velocity at the point of impact at ground level.

d) If Vx and Vy are the horizontal and vertical components of a vector, then:

|V| = sqrt(Vx^2 + Vy^2)

so we can compute the magnitude of the impact velocity

e) As above, if the total velocity is V and the angle is X:

horizontal component = V x cos X
vertical component = V x sin X

so (vertical V)/(horizontal V) = (sin X)/(cos X) = tan X
So the angle is the arc tangent of the velocity ratio. Or you could use arc sine or arc cosine with the other ratios.

5) We already did this one.

Answer 2

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