Projectile motion. pls help?

Question

1. A projectile is shot (canonball) from the edge of a cliff 125m above the ground level with an initial speed of 65.0m/s at an angle of 37.0º with the horizontal.

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

(b) Determine the range of the projectile as measured from the base of the cliff. At the instant just before the projectile hits the ground level.

(c) The horizontal and vertical components of its velocity.

(d) Magnitude of the velocity.

(e) 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

1. First use the angle to separate the initial velocity into two components, the horizontal and the vertical velocities.

2. Using the initial vertical velocity and the velocity equation for motion under constant linear acceleration:

Vf = Vi + A T

determine the time the projectile rises (i.e. until its vertical velocity is 0) using the acceleration due to gravity, g

3. Using the distance equation for motion under constant linear acceleration:

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

determine the height above the cliff that the projectile reaches.

4. Using the distance equation, determine the time  the projectile takes to fall to ground level. The total travel time is time up + time down.

5. The total horizontal distance traveled is total time x horizontal velocity. (The horizontal velocity stays constant throughout)

6. Using the velocity equation and the falling time, determine the final vertical velocity of the projectile.

7. Combine the horizontal and vertical components of the final velocity to compute the magnitude and angle of the velocity on impact.

Answer 2

At time t after launch, the projectile's co-ordinates measured from ground level are:

x = 65.0t cos(37.0) ...(1)

y = 125 + 65.0t sin(37.0) - 9.81t^2 / 2 ...(2)

Putting y = 0 and solving (2) for t:

9.81t^2 / 2 - 65.0 t sin(37.0) - 125 = 0

t = (65.0 sin(37.0) +/- sqrt(65.0^2 sin^2(37.0) + 2 * 9.81 * 125)) / 9.81

= 10.4 sec.

(b)

Substituting this value of t in (1):

x = 541 m.

(c)

x' = 65.0 cos(37)

= 51.9 m/s.

y' = 65.0 sin(37) - 9.81 * 10.4

= - 63.1 m/s. upward

= 63.1 m/s downward.

(d)

sqrt(51.9^2 + 63.1^2)

= 81.6 m/s.

(e)

If a is the angle,

tan(a) = y' / x'

a = 50.6 deg.

(f)

If h is the height:

0 = 65.0^2 sin^2(37.0) - 2 * 9.81h

h = 78.0 m.