What is the height of the cliff

Question

A ballis launched at a bottom of a cliff with initial speed v0=40m/s directed at an angle of theta= 42° above the horizontal towards a cliff with height x. The ball hits the cliff 4.50s after the launch. Ffind the following:

a. Height x

b. horizontal displacement of ball

C. Maximum height reached by the ball

D. Speed of ball before the impact

Answer 1

If v₀ = 40 m/s at an angle of 42° from the horizontal, then we can use trig ratios to find the components of v₀ in the x-direction (v_0-x) and in the y-direction (v_0-y):

cos(42°) = adj/hyp = v_0-x/40 meaning v_0-x = 40cos(42°) = 29.726 m/s

sin(42°) = opp/hyp = v_0-y/40 meaning v_0-y = 40sin(42°) = 26.765 m/s

The only force on the ball as it is flying though the air is the force of gravity which acts in the downward direction. That means in the x-direction, there is no force acting to change the velocity. Consequently, IN THE X-DIRECTION, the velocity remains the same as v_0-x meaning the velocity in the x-direction at all locations and times remains at 29.726 m/s until it hits the cliff. Since it hits the cliff in 4.50 s then the distance to the cliff is (29.726 m/s)(4.50 s) = 133.766 meters = 134 m. This is the "horizontal displacement of the ball".

In the y-direction, the travel acts just like it were traveling straight up and then free-falling back down. To find the maximum height of the ball, we can use the kinematic equation of motion:

v_f-y² = v_0-y² + 2a(Δy) where v_f-y = 0 (peak height of the ball), v_0-y = 26.765 m/s, a = -9.81 m/s²:

0² = 26.765² + 2(-9.81)(Δy)

0 = 716.377 - 19.62(Δy)

Δy = 36.513 meters = 36.5 m (maximum height reached by the ball)

Again, in the y-direction, the travel acts just like it were traveling straight up and then free-falling back down. To find the height "x" of the ball, we can use the kinematic equation of motion:

x_f = x₀ + v₀t + 1/2at² where x_f = "x", x₀ = 0, v0 = v_0-y = 26.765 m/s, t = 4.50 s, a = -9.81 m/s²:

x = 0 + (26.765)(4.5) + 1/2(-9.81)(4.50²)

x = 21.117 = 21.1 m

To find the velocity at the time the ball hits the cliff, we will need to know the two components of the velocity, v_f-x and v_f-y. Remember that v-x remains constant at 29.726 m/s so v_f-x = 29.726 m/s. To find vf-y, use the kinematic equation of motion:

v_f-y² = v_0-y² + 2a(Δy) where Δy = our "x":

v_f-y² = 26.765² + 2(-9.81)(21.117)

v_f-y² = 716.377 - 414.321

v_f-y = 17.3798 m/s = 17.4 m/s

The direction of travel is downward. The reason can be seen by solving the kinematic equation of motion:

x_f = x₀ + v₀t + 1/2at² to find the values of time for which the height is "x":

21.117 = 0 + 26.765t + 2(-9.81)t²

-4.905t² + 26.765t - 21.117 = 0

Solving using the quadratic formula yields t = 0.957 s and t = 4.500 s so we see that the ball passes x = 21.1 meters at t = 0.957 s going up and then reaches 21.1 meters at t = 4.50 s going down.

To find the velocity at this time, we need to combine v_f-x and v_f-y which is achieved by getting the magnitude from the Pythagorean Theorem: v_√f = √(v_f-x² + v_f-y²) = √(29.726² + (-17.3798)²) = √1185.679 = 34.4 m/s. The direction is attained by using inverse tangent = tan^-1(-17.3798/29.726) = -30.3°

The final velocity is 34.4 m/s at an angle of 30.3° below the horizontal.

Answer 2

I will give you the appropriate equations to solve:

a) y(t = 4.5s) = v_y0t - (1/2)gt² where v_y0 = 40 m/s sin(42) and g = 9.8 m/s²

b) x(t = 4.5 s) = v_x0t_air = (40 m/s)cos(42) * 4.5 s

c) y_max = v_y0²/(2g)

d) v_y(t) = v_y0 - gt and v(t) = sqrt(v_y² + v_x0²)

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