You throw a ball downward from a window at a speed of 2.7 m/s. How fast will it be moving when it hits the sidewalk 2.8 m below?
How far does a plane fly in 13 s while its velocity is changing from 142 m/s to 75 m/s at a uniform rate of acceleration?
You throw a ball upward from a window at a speed of 1.7 m/s. How fast will it be moving when it hits the sidewalk 2.5 m below?
To determine how fast the ball will be moving when it hits the sidewalk, we can use the equations of motion.
First, let's define our variables:
- Initial velocity (u) = 2.7 m/s (downward direction)
- Final velocity (v) = ? (we need to find this)
- Distance (s) = 2.8 m (downward direction)
- Acceleration (a) = 9.8 m/s^2 (acceleration due to gravity acting downwards)
Now, we can use the equation of motion that relates final velocity, initial velocity, acceleration, and distance:
v^2 = u^2 + 2as
Plugging in the values, we get:
v^2 = (2.7 m/s)^2 + 2 * (-9.8 m/s^2) * (2.8 m)
v^2 = 7.29 m^2/s^2 - 54.88 m^2/s^2
v^2 = -47.59 m^2/s^2
Since the value inside the square root is negative, it means that there is no real solution for v^2. This indicates that the ball will not hit the sidewalk because its initial downward speed is not high enough. It will reach a maximum height and then fall back towards the window.