A stone is thrown with an initial speed of 15 m/s at an angle of 53° above the horizontal from the top of a 40-m building.If g = 9.8 m/s² and air resistance is negligible,then what is the magnitude of the horizontal component of velocity as the rock strikes the ground? A) 7.5 m/s B) 9.0 m/s C) 12 m/s D) 29 m/s

Question

Quizplus · Accepted Answer

Using the equations of motion, we can find the time taken for the stone to hit the ground:
S = ut + 1/2 at^2
Here, initial vertical velocity (u) = 15 sin 53° = 11.7 m/s (upwards)
a = g = 9.8 m/s^2 (downwards)
S = 40 m (height of the building)
Putting the values, we get:
40 = 11.7t - 1/2 (9.8)t^2
Solving this quadratic equation, we get t ≈ 3.67 s as the time taken.
Now, we can find the final horizontal velocity (v) using:
v = u + at
Here, initial horizontal velocity (u) = 15 cos 53° = 9.65 m/s
a = 0 (horizontal acceleration is 0)
t = 3.67 s (time taken)
Therefore,
v = u + at = 9.65 m/s + 0 × 3.67 s = 9.65 m/s

Hence, the answer is B.

A Stone Is Thrown with an Initial Speed of 15