a ball is rolled off the edge of a dining table 1.5 m high with a horizontal velocity of 2.2 m/s with what velocity does it strike the ground? How far from the edge of the table did it land on the floor?
What is the answer
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To determine the velocity of the ball when it strikes the ground and the horizontal distance it travels before landing, we can use the principles of projectile motion.
First, let's calculate the time it takes for the ball to fall from the table to the ground. We can use the equation for vertical motion:
h = 0.5 * g * t^2
Where:
h = height (1.5 m)
g = acceleration due to gravity (9.8 m/s^2)
t = time
Rearranging the equation to solve for time:
t = sqrt(2h/g)
Plugging in the values:
t = sqrt(2 * 1.5 / 9.8) ≈ 0.55 s
Now, let's calculate the vertical component of the ball's velocity when it hits the ground. We can use the equation:
v_vertical = g * t
Plugging in the values:
v_vertical = 9.8 * 0.55 ≈ 5.39 m/s
The ball's vertical velocity when it hits the ground is approximately 5.39 m/s. However, since the ball is rolling off the table horizontally, its vertical velocity is purely due to gravity. So the horizontal component of the ball's velocity will remain constant at 2.2 m/s.
To find the magnitude of the final velocity, we can use the Pythagorean theorem:
v_final = sqrt(v_horizontal^2 + v_vertical^2)
Plugging in the values:
v_final = sqrt(2.2^2 + 5.39^2) ≈ 5.89 m/s
Therefore, the ball strikes the ground with a velocity of approximately 5.89 m/s.
Next, let's calculate the horizontal distance the ball travels before landing on the floor. We can use the equation:
d = v_horizontal * t
Plugging in the values:
d = 2.2 * 0.55 ≈ 1.21 m
Therefore, the ball lands on the floor at a horizontal distance of approximately 1.21 m from the edge of the table.
You can put these in Physics. Anyway...
t = sqrt(2*1.5/9.8)
x = 2.2t
for velocity
vx is constant at 2.2
vy = 9.8t
Use Pythagorean to find total v