A ball is thrown horizontally from the roof of a building 59.0 tall and lands 27.4 from the base. What was the ball's initial speed?
To find the initial speed of the ball, we can use the fact that the horizontal distance traveled by the ball is equal to the initial horizontal speed of the ball multiplied by the time of flight.
First, let's find the time of flight of the ball. We know that the ball was thrown horizontally, which means that the vertical motion of the ball is only due to the force of gravity. We can use the equation of motion in the vertical direction:
Δy = v₀y * t + (1/2)*g*t²
Where:
Δy is the vertical displacement (the height of the building, which is 59.0 m),
v₀y is the initial vertical velocity (which is 0, since the ball is thrown horizontally),
g is the acceleration due to gravity (which is approximately 9.8 m/s²),
t is the time of flight.
Plugging in the values, the equation becomes:
59.0 = 0 * t + (1/2)*9.8*t²
Simplifying the equation:
29.4t² = 59.0
Dividing both sides by 29.4:
t² = 2.0
Taking the square root of both sides:
t = √(2.0)
t ≈ 1.41 s
Now that we know the time of flight, we can use the horizontal distance traveled by the ball to find the initial horizontal speed. The equation is:
Δx = v₀x * t
Where:
Δx is the horizontal distance traveled by the ball (27.4 m),
v₀x is the initial horizontal velocity (the value we want to find),
t is the time of flight (1.41 s).
Plugging in the values, the equation becomes:
27.4 = v₀x * 1.41
Solving for v₀x:
v₀x = 27.4 / 1.41
v₀x ≈ 19.43 m/s
Therefore, the ball's initial speed was approximately 19.43 m/s.