A man standing on a bridge throws a stone horizontally with a speed of 20 m/s. The stone hits the water below 2.5 s later. The bridge is
A) 44.1 m high.
B) 60.1 m high.
C) 30.6 m high.
D) 20.3 m high.

Question

Quizplus · Accepted Answer

We can use the equation:
y = vi*t + (1/2)*a*t^2
where y is the vertical displacement, vi is the initial vertical velocity (which is 0 m/s since the stone is thrown horizontally), a is the acceleration due to gravity (-9.8 m/s^2), and t is the time of flight (2.5 s).

We want to solve for the height of the bridge, so we need to find the vertical displacement. Since the stone is thrown horizontally, the horizontal displacement is given by:
x = vi*x*t
where vi,x is the initial horizontal velocity (20 m/s) and x is the horizontal distance traveled (which we don't know yet).

We can rearrange this equation to solve for x:
x = vi,x*t
x = 20 m/s * 2.5 s
x = 50 m

Now we can use the vertical displacement equation with our values:
y = vi*t + (1/2)*a*t^2
y = 0 m/s * 2.5 s + (1/2)*(-9.8 m/s^2)*(2.5 s)^2
y = -30.625 m

Since the stone falls below the level of the bridge, the height of the bridge is equal to the absolute value of the vertical displacement:
|y| = 30.625 m

Therefore, the closest answer choice is C) 30.6 m high.

A Man Standing on a Bridge Throws a Stone Horizontally