A ball is thrown from a cliff upward with a speed of 21 m/s and at an angle of 56 degrees to the horizontal. How high does it go above the cliff to the nearest tenth of a meter?
Initial vertical velocity component = Voy = 21 sin 56 = 17.41 m/s
Voy^2 = 2 g H
Solve for H
To find the maximum height reached by the ball, we first need to calculate the vertical component of its initial velocity. Since the ball is thrown upward at an angle of 56 degrees to the horizontal, the vertical component can be calculated using the formula:
Vertical Velocity (Vy) = Initial Velocity (V) * sin(angle)
Given that the initial velocity (V) is 21 m/s and the angle is 56 degrees, we can substitute these values into the formula:
Vy = 21 m/s * sin(56°)
Now we need to calculate the time it takes for the ball to reach its highest point. To do this, we can use the formula:
Time (t) = (Vertical Velocity (Vy)) / (Gravity (g))
Where gravity (g) is a constant value of approximately 9.8 m/s^2.
Using the given values, we can calculate the time:
t = Vy / g
= (21 m/s * sin(56°)) / 9.8 m/s^2
Once we have the time it takes for the ball to reach the highest point, we can use it to find the maximum height reached by the ball. The formula to calculate the maximum height (h) is:
h = (Vertical Velocity (Vy)^2) / (2 * Gravity (g))
Substituting the values we calculated earlier:
h = (Vy^2) / (2 * g)
= ((21 m/s * sin(56°))^2) / (2 * 9.8 m/s^2)
Evaluating this expression will give us the maximum height of the ball above the cliff.