Consider a rock thrown off a bridge at a height of 78.1 m that is thrown at an a

Question

Consider a rock thrown off a bridge at a height of 78.1 m that is thrown at an angle of 25* with respect to the horizontal x axis. The initial speed of the rock is 14. M/s find the following
The maximum height reached by the rock
The time it takes the rock to reach its maximum height
The Time at which the rock lands
How far horizontally from the bridge the rock lands
The velocity of the rock (magnitude and direction) just before it lands

Potentially Useful Equations net mã FA on B=-FB on A 1 .IFNI u2 F,-(mg, downward) 151 mu2 Potentially Useful Equations From Exam 1 Ar g = 9.8 m/? IF,I = mg 1 m/s2.2 mph I mi 1.609 km

Explanation / Answer

i) using uy2 = uyo2 - 2gh, we get

0 = (14 sin25)2 - 2 *(9.8)* h

Therefore, h = (14 sin25)2 / 2(9.8) = 1.786 m

adding that to 78.1m ( the bridge's height) we get :

hmax = 78.1 + 1.786 = 79.886 m

ii) at max height uy = 0

so , using uy = uyo - gt

=> 0 = 14sin25 - 9.8t

=> t = 14sin25 / 9.8 = 0.604 s (approx)

iii) when the rock lands then h = - 78.1 m

so - 78.1 = 14sin25 t - 0.5(9.8) t2

=> -78.1 = 5.92 t - 4.9 t2

=> 4.9 t2 - 5.92 t - 78.1 = 0

t = [5.92 + sqrt (5.382 + 4(4.9)(78.1))] / 2(4.9) = 4.64s

iv) ux remains consant , so we use ux = x / t

=> x = ux t = 14 cos25 x 4.64 = 58.87 m away from the bridge

v) ux = 14 cos 25 = 12.69 m/s

now find uy :

uy = 14sin25 - 9.8 (4.64) = - 39.55 m/s

so V = sqrt(12.692 + 39.552) = 41.53 m/s

angle = arctan (-39.55 / 12.69) = - 72.21 degrees

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