A ball is thrown horizontally from the top of
a building 130 m high. The ball strikes the
ground 70 m horizontally from the point of
release.
What is the speed of the ball just before it
strikes the ground?
Answer in units of m/s.
To find the speed of the ball just before it strikes the ground, we can use the kinematic equations of motion. Since the ball is thrown horizontally, the initial vertical velocity is 0 m/s.
Given:
Initial height, h = 130 m
Horizontal distance, x = 70 m
Initial vertical velocity, uy = 0 m/s
Final vertical velocity, vy = ? (since the ball is about to hit the ground)
Acceleration due to gravity, g = 9.8 m/s²
We can use the kinematic equation for vertical motion:
vy² = uy² + 2gΔy
where Δy is the change in height. In this case, Δy = -h, as the ball is falling downwards.
Plugging in the values we have:
vy² = 0² + 2 * 9.8 * (-130)
vy² = - 2 * 9.8 * 130
vy² = -2548
Since the square of a velocity cannot be negative, we ignore the negative sign:
vy = √2548
vy ≈ 50.48 m/s
Therefore, the speed of the ball just before it strikes the ground is approximately 50.48 m/s.
To find the speed of the ball just before it strikes the ground, we can use the equations of motion.
We know that the ball was thrown horizontally, so its initial vertical velocity (Vy) is zero. The only force acting on the ball is gravity, which causes it to accelerate vertically downward.
First, let's find the time it takes for the ball to reach the ground using the equation:
h = (1/2) * g * t^2
where h is the height of the building (130 m), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time.
Rearranging the equation, we get:
t = sqrt(2h/g)
Substituting the given values, we get:
t = sqrt(2 * 130 / 9.8) = sqrt(26) ≈ 5.1 seconds
Now, let's find the horizontal velocity (Vx) of the ball. Since there are no horizontal forces acting on the ball, its velocity remains constant throughout its motion. The horizontal distance traveled by the ball is 70 m, and the time it takes is 5.1 seconds, so we can calculate the horizontal velocity using:
Vx = d / t
where d is the horizontal distance and t is the time.
Substituting the given values, we get:
Vx = 70 / 5.1 ≈ 13.725 m/s
Finally, to find the speed of the ball just before it strikes the ground, we can use the Pythagorean theorem:
Speed = sqrt(Vx^2 + Vy^2)
Since Vy is zero, the equation simplifies to:
Speed = sqrt(Vx^2) = Vx ≈ 13.725 m/s
Therefore, the speed of the ball just before it strikes the ground is approximately 13.725 m/s.
V^2 = Vo^2 + 2g*h. = 0 + 19.6*130 = 2548,
V =