A ball is thrown horizontally from the top of
a building 37.5 m high. The ball strikes the
ground at a point 80.3 m from the base of the
building.
The acceleration of gravity is 9.8 m/s
2
.
Find the time the ball is in motion.
Answer in units of s.
2.77 s
To find the time the ball is in motion, we can use the equations of motion and apply them to the horizontal motion.
Given:
Initial vertical position (y₀) = 37.5 m
Initial vertical velocity (v₀) = 0 m/s (since the ball is thrown horizontally)
Acceleration due to gravity (g) = 9.8 m/s²
Since the ball is thrown horizontally, the horizontal component of its velocity remains constant throughout its motion, and it covers a horizontal distance of 80.3 m.
We can use the equation for horizontal motion to find the time (t) it takes for the ball to cover the horizontal distance:
Distance (d) = Velocity (v) x Time (t)
d = v₀ * t
Since the horizontal velocity remains constant, the initial velocity in the horizontal direction is equal to the final velocity. Therefore, we can also write:
d = (v₀ + v) / 2 * t
d = (0 + v) / 2 * t
d = v/2 * t
Rearranging the equation, we get:
t = (2 * d) / v
Substituting the given values, we have:
d = 80.3 m
Now, we need to find the horizontal component of the velocity (v). Since there is no acceleration in the horizontal direction, the horizontal velocity remains constant. We can use the equation:
v = d / t
Substituting the given values, we have:
d = 80.3 m
t = ?
v = ?
Now, rearrange the equation to solve for v:
v = d / t
v = v₀
Substituting the given values, we have:
v = 80.3 m / t
Now, equate the two equations for v:
80.3 m / t = v
Substitute this value of v into the equation for time:
t = (2 * d) / (80.3 m / t)
Simplify:
t² = (2 * d) / (80.3)
t² = (2 * 80.3 m) / (80.3)
t² = 2
t = sqrt(2)
Therefore, the time the ball is in motion is approximately equal to the square root of 2 seconds (s).