A ball thrown horizontally from the top of a 13m high building strikes the ground 5m from the base of the building.
1) With what velocity was the ball thrown?
To determine the velocity at which the ball was thrown, we can use the concept of projectile motion.
First, let's analyze the initial horizontal motion of the ball. Since the ball was thrown horizontally, the initial vertical velocity is 0 m/s. The horizontal distance traveled by the ball is 5m.
Now, let's focus on the vertical motion of the ball. The initial vertical position is 13m above the ground, and the final vertical position is at ground level, which means the vertical displacement is -13m (negative because it is going downward). The acceleration due to gravity is -9.8 m/s^2 (accelerating downward).
We can use the following kinematic equation to relate the variables:
Δy = Vy * t + (1/2) * a * t^2
In this case, Δy is -13m (negative because it is going downward), Vy is 0 m/s, and a is -9.8 m/s^2. We need to solve for t (time of flight).
Plugging in the values and rearranging the equation, we have:
-13 = 0 * t + (1/2) * (-9.8) * t^2
-13 = -4.9 * t^2
Dividing both sides by -4.9, we get:
t^2 = 2.653
t ≈ √2.653 ≈ 1.63 seconds
Now that we have the time of flight, we can find the horizontal velocity using the equation:
Vx = Δx / t
Plugging in Δx = 5m and t ≈ 1.63s into the equation, we have:
Vx = 5 / 1.63 ≈ 3.07 m/s
Therefore, the ball was thrown horizontally with a velocity of approximately 3.07 m/s.
find the time it takes to fall 13m.
then, v=5/time