Consider a rock thrown off a bridge of height 73.6 m at an angle theta = 25 degr

Question

Consider a rock thrown off a bridge of height 73.6 m at an angle theta = 25 degree with respect to the horizontal as shown in Figure P4.20. The initial speed of the rock is 16.1 m/s. Find the following quantities: 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. 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 sign!)

Explanation / Answer

a) The initial vertical velocity of the rock
v0v  = v0 sin 250
v0v = 16.1 m/s x sin 25 = 6.80 m/s
The rock attains its maximum height when the vertical velocity becomes zero
v2  = u2 + 2 a S
Where S is the distance traveled , v is the final velocity, u is the initial velocity and a is the acceleration
0 = v0v2 - 2 g S
S = v0v2 / 2 g
S = 6.802 / 2 x 9.81 = 46.29 m
The height measured from the bottom is
H =46.29 m + 73.6 m = 119.89 m
b) Using the equation
v = u + a t
0 = v0v - g t
t = v0v / g
t = 0.693 s
c) After reaching the maximum height, the rock is in free fall motion, the time taken by the rock to reach the bottom from the maximum height , from the equation
H = (1/2) g t2
t = sqrt (2 H / g)
t = sqrt (2 x 119.89 / 9.81)
t = 4.94 s
The total time taken by the rock is
t = 4.94 s + 0.693 s = 5.64 s
d) The horizontal distance traveled is
H = vh t
Where vh is the horizontal velocity and t is the total time taken
vh  = v0 cos 250  = 16.1 m/s x cos 25 = 14.59 m/s
H = 14.59 m/s x 5.64 s = 82.3 m

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