Consider a rock thrown off a bridge of height 77.3 m at an angle = 25° with resp

Question

Consider a rock thrown off a bridge of height 77.3 m at an angle = 25° with respect to the horizontal as shown in Figure P4.20. The initial speed of the rock is 18.2 m/s. Find the following quantities:

(a) the maximum height reached by the rock

(b) the time it takes the rock to reach its maximum height

(c) the time at which the rock lands

(d) how far horizontally from the bridge the rock lands

(e) the velocity of the rock (magnitude and direction) just before it lands.

magnitude

(f) direction: Give your answer in degrees. If you think the answer is "20 degrees down from the positive x-axis", you would enter "-20" (note the negative sign!)

Explanation / Answer

a) at max height Vy = 0

so , using Vy = Vyo - gt

=> 0 = 18.2*sin25 - 9.8*t

=> t = 18.2*sin25 / 9.8 = 0.785 = 0.78 s (approx)

b) now using Vy^2 = Vyo^2 - 2gh

0 = (18.2*sin25)^2 - 2 (9.8)*h

h = (18.2*sin25)^2 / (2*9.8) = 3.02 m

adding that to 77.3 m ( the bridge's height) we get

h(max) = 77.3 + 3.02 = 80.32 m

c) when the rock lands then h = - 80.32 m

using equation S = uy*t + (1/2)ay*t^2

so - 80.32 = 18.2*sin25*t - 0.5*(9.8)*t^2

=> - 80.32 = 7.69*t - 4.9*t^2

=> 4.9*t^2 - 7.69*t - 80.32 = 0

t = [7.69 + sqrt (7.69^2 + 4*(4.9)*(80.32))] / (2*4.9) = 4.9 s

d) Vx remains consant , so we use Vx = x / t

=> x = Vx*t = 18.2*sin25*4.9 = 37.68 m away from the bridge

e) Vx = 18.2*sin25 = 7.69 m/s

now find Vy :

Vy = 18.2*sin25 - 9.8*(4.9) = - 40.33 m/s

so V = sqrt(7.69^2 + 40.33^2) = 41.05 m/s

angle = arctan (- 40.33 / 7.69) = - 79.2 degrees { down from the positive x-axis }

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