Consider a rock thrown off a bridge of height 80 m at an angle = 25° with respec

Question

Consider a rock thrown off a bridge of height 80 m at an angle = 25° with respect to the horizontal as shown in Figure P4.20. The initial speed of the rock is 19 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

0.819 s

You are correct.(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

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) maximum height above ground = 80+ (u sin theta) ^2/2g

= 80 + (19*sin 25 degree) ^2 /(9.8*2)

= 83.3 m

B) T = usin theta/g = 19 sin 25 degree /9.8

= 0.819 s

D) total time T = 0.819 + sqrt(2 h/g) = 0.819 + sqrt(2*83.3/9.8)

= 4.94 s

Horizontal distance = v cos theta T

= 19 cos 25 degree *4.94

= 85.07m

E) Vx = 19* cos 25 degree = 17.22 m/s

Vy = - gt = - 9.8*4.94 = - 48.41 m/s

Magnitude = sqrt(17.22^2 +48.41^2)

= 51.4 m/s

Direction = arctan (-48.41/17.22)

= - 70.4 degree

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