A shoe flung into the air such that at the end of 2.0 m/s it is at its maximum height moving at 6.0 m/s. How far away will it be from the thrower when it returns to the launch altitude.
2.0 what?
2.0 s
by the way at max height, your entire velocity is the horizontal component, u = 6 m/s , which is constant until it hits something.
so it took 2 seconds to stop vertical motion
so it was in the air for 4 seconds
so it went 6 * 4 = 24 meters
To solve this problem, we can use the concept of projectile motion. Let's break down the given information:
Initial velocity (u) = 2.0 m/s (upward)
Final velocity (v) = 6.0 m/s (downward)
Acceleration (a) = -9.8 m/s² (due to gravity, always directed downwards)
Distance traveled upwards = Distance traveled downwards (since the shoe returns to the launch altitude)
We need to find the distance traveled when the shoe returns to the launch altitude. To do this, we'll make use of the following equation:
v² = u² + 2ad
Where:
v = final velocity
u = initial velocity
a = acceleration
d = distance
Since the initial velocity is upwards, we can assign it a negative value to maintain consistent signs with the acceleration. Therefore:
u = -2.0 m/s
Simplifying the equation to solve for distance (d), we have:
d = (v² - u²) / (2a)
Substituting the values we have:
d = (6.0² - (-2.0)²) / (2 * -9.8)
Calculating this expression gives us:
d = 19.8 / -19.6
d ≈ -1.01 m
The negative sign indicates that the distance is in the opposite direction of the shoe's launch. So, the shoe will be approximately 1.01 meters away from the thrower when it returns to the launch altitude.