A ball is thrown vertically upward with a speed of 15.0 m/s.
(a) How high does it rise?
(b) How long does it take to reach its highest point?
(c) How long does the ball take to hit the ground after it reaches its highest point?
(d) What is its velocity when it returns to the level from which it started?
1.53
To find the answers to these questions, we can use the equations of motion for projectile motion and the principles of free fall.
(a) To calculate how high the ball rises, we can use the equation:
Δy = v₀t - 0.5gt²
Where:
Δy = vertical displacement (height)
v₀ = initial velocity (15.0 m/s)
t = time taken to reach the highest point (unknown)
g = acceleration due to gravity (approximately 9.8 m/s²)
Since the ball is thrown vertically upward, it will eventually reach a point where its final velocity becomes zero. At that point, it will start falling down.
Using the equation, we can find the height:
0 = 15.0t - 0.5 * 9.8 * t²
Rearranging the equation to get it in the form of a quadratic equation:
4.9t² - 15t = 0
t(4.9t - 15) = 0
This equation has two possible solutions: t = 0 and 4.9t - 15 = 0
Since time cannot be negative, the solution is t = 0.
Therefore, the ball reaches its highest point instantaneously, and the height it rises is 0.
(b) The time it takes to reach the highest point can be calculated using the equation:
v = v₀ - gt
At the highest point, the final velocity (v) is zero, so:
0 = 15.0 - 9.8t
Solve for t:
9.8t = 15.0
t = 15.0 / 9.8
Therefore, the time taken to reach the highest point is approximately 1.53 seconds.
(c) The time it takes for the ball to hit the ground after reaching its highest point is the same as the time it took to reach the highest point, which is approximately 1.53 seconds.
(d) When the ball returns to the level from which it started, its velocity will be the negative of its initial velocity because it's moving in the opposite direction. Therefore, the velocity when it returns to the initial level is -15.0 m/s.