A ball with a horizontal speed of 1.5 m/s rolls off a bench that is 2.2 m high. How long will it take to reach the floor and how far will the ball land away from the bench?
4.9t^2 = 2.2
find t, and then the horizontal distance is
d = 1.5t
To find the time it takes for the ball to reach the floor, we can use the equation of motion:
\[ h(t) = h_0 + v_0t + \frac{1}{2}gt^2 \]
Where:
- \( h(t) \) is the height of the ball at time \( t \)
- \( h_0 \) is the initial height (2.2 m)
- \( v_0 \) is the initial vertical velocity (0 m/s since the ball only has horizontal velocity)
- \( g \) is the acceleration due to gravity (-9.8 m/s²)
At the moment the ball rolls off the bench, its height is \( h(t) = h_0 \). When the ball reaches the ground, its height \( h(t) = 0 \). Substituting these values into the equation, we get:
\[ 0 = h_0 + v_0t + \frac{1}{2}gt^2 \]
Simplifying the equation, we now have a quadratic equation in terms of \( t \):
\[ 0 = 2.2 + 0.5 \times (-9.8) \times t^2 \]
Solving for \( t \):
\[ 0 = 2.2 - 4.9t^2 \]
\[ 4.9t^2 = 2.2 \]
\[ t^2 = \frac{2.2}{4.9} \]
\[ t = \sqrt{\frac{2.2}{4.9}} \]
By calculating the square root, we find the value of \( t \).
To find the time it takes for the ball to reach the floor, we can use the equation of motion:
h = ut + (1/2)gt^2
Where:
h = height (2.2 m)
u = initial vertical velocity (0 m/s)
g = acceleration due to gravity (-9.8 m/s^2)
Since the ball rolls off horizontally and does not have an initial vertical velocity, u = 0 m/s. Let's calculate the time it takes for the ball to reach the floor:
2.2 = 0t + (1/2)(-9.8)t^2
Rearranging the equation:
4.9t^2 = 2.2
Dividing both sides by 4.9:
t^2 = 2.2 / 4.9
Taking the square root of both sides:
t ≈ √(2.2 / 4.9) ≈ 0.48 seconds
Therefore, it will take approximately 0.48 seconds for the ball to reach the floor.
To find the horizontal distance the ball will land away from the bench, we can use the formula:
d = ut
Where:
d = horizontal distance
u = horizontal velocity (1.5 m/s)
t = time (0.48 seconds)
Plugging in the values:
d = 1.5 * 0.48 ≈ 0.72 meters
Therefore, the ball will land approximately 0.72 meters away from the bench.