Projectile Motion shot over a cliff

[sw3etazngyrl]

1. A projectile is shot from the edge of a cliff 125m above ground level with an initial speed of 105m/s at an angle of 37.0 degrees with the horizontal. a) Determine time taken by projectile to hit point P at ground level. b) Determine range X of the projectile as measured from the base of cliff. At the instant just before the projectile hits point P, find c) the horizonal and vertical components of its velocity, d) the magnitude of the velocity, and e) the angle made by the velocity vector with the horizontal.

2. For c, how do you find the vertical/horizontal components? Should I use my answer for a to complete part c?

3. a) t=sqrt(2h/g)=sqrt(2*125m/9.81m/s^2)=5.05 s
b) x=105m/s * 5.05s=530 m

that's all i have so far. i don't know how to start c and the rest after that.

 

[Astronuc]

Refer to this -

http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra9

and in general -

http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html

 

[m2003]

For part c, you can use the equations for horizontal and vertical components of velocity:

Horizontal component of velocity (Vx) = V0 * cos(theta)
Vertical component of velocity (Vy) = V0 * sin(theta)

Where V0 is the initial velocity and theta is the angle of projection.

Using the values given in the problem, we can calculate the horizontal and vertical components of velocity:

Vx = (105m/s) * cos(37.0 degrees) = 84.0 m/s
Vy = (105m/s) * sin(37.0 degrees) = 63.9 m/s

For part d, we can use the Pythagorean theorem to find the magnitude of the velocity (V):

V = sqrt(Vx^2 + Vy^2) = sqrt((84.0 m/s)^2 + (63.9 m/s)^2) = 106 m/s

For part e, we can use the inverse tangent function to find the angle made by the velocity vector with the horizontal:

Angle = tan^-1(Vy/Vx) = tan^-1(63.9 m/s / 84.0 m/s) = 36.9 degrees

You can use your answer for part a to complete part c, as the time taken by the projectile to hit point P is the same as the time taken for the projectile to reach the horizontal distance (x) from the base of the cliff.