A basketball player is running at 5.00 m/s directly toward the basket when he jumps into the air to dunk the ball. He maintains his horizontal velocity.
(a) What vertical velocity does he need to rise 0.750 meters above the floor?
(b) How far from the basket (measured in the horizontal direction) must he start his jump to reach his maximum height at the same time as he reaches the basket?
a. Y^2 = Yo^2 + 2g*h = 0 m/s @ max. ht.
Yo^2 - 19.6*0.750 = 0
Yo^2 = 14.7
Yo = 3.83 m/s. = Initial ver. velocity.
b. Y = Yo + g*Tr = 0 m/s @ max. ht.
Y = 3.83 - 9.8Tr = 0
9.8Tr = 3.83
Tr = 0.3908 s. = Rise time.
Dx=Xo * Tr = 5m/s * 0.3908s. = 1.95 m.
From basket.
To solve this problem, we need to consider the motion of the basketball player in both the vertical and horizontal directions.
(a) To find the vertical velocity needed to rise 0.750 meters above the floor, we can use the kinematic equation:
Δy = v_iy * t + (1/2) * a_y * t^2
where
Δy = vertical displacement (0.750 m),
v_iy = initial vertical velocity (to be determined),
t = time taken to reach the maximum height (unknown),
a_y = vertical acceleration (due to gravity, approximately -9.8 m/s^2).
Since the player maintains his horizontal velocity throughout the jump, there is no horizontal acceleration. Therefore, the time taken to reach the maximum height will be the same as the time taken to reach the basket.
Simplifying the equation, we get:
0.750 m = v_iy * t + (1/2) * (-9.8 m/s^2) * t^2
Now, we need to determine the time taken to reach the maximum height. The vertical velocity component at the maximum height is zero (since the player reaches the highest point of their jump). So, we can use the equation:
v_fy = v_iy + a_y * t
Since v_fy = 0 at the maximum height, we have:
0 = v_iy + (-9.8 m/s^2) * t
We can solve this equation for t and substitute it into the previous equation to find v_iy.
0 = v_iy - 9.8t --------------(1)
v_iy*t + (1/2) * (-9.8) * t^2 = 0.750 ----------------(2)
From equation (1), v_iy = 9.8t $$$$$$ -------------(3)
Substitute equation (3) into equation (2):
(9.8t)*t + (1/2) * (-9.8) * t^2 = 0.750
Simplifying, we get:
4.9t^2 - 0.5t + 0.750 = 0
Now we need to solve this quadratic equation for t. We can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
where a = 4.9, b = -0.5, and c = 0.750.
Using the quadratic formula, we find two possible values for t. Choose the positive value, as negative time is not meaningful in this context.
t ≈ 0.39 s
Now that we have the value of t, we can substitute it back into equation (3) to find v_iy:
v_iy = 9.8t
v_iy ≈ 9.8 * 0.39
v_iy ≈ 3.82 m/s
Therefore, the player needs a vertical velocity of approximately 3.82 m/s to rise 0.750 meters above the floor.
(b) To find the horizontal distance from the basket where the player must start their jump to reach the maximum height at the same time as reaching the basket, we can use the horizontal velocity:
v_ix = 5.00 m/s
Since there is no horizontal acceleration, the horizontal velocity remains constant throughout the jump.
Now we need to determine the time taken to reach the basket. We can use the equation:
Δx = v_ix * t
where Δx is the horizontal distance from the basket.
Rearranging the equation to solve for t, we get:
t = Δx / v_ix
Since we want the player to reach their maximum height at the same time as reaching the basket, the time taken to reach the maximum height is the same as the time taken to reach the basket:
t ≈ 0.39 s (as calculated earlier)
Substituting this value of t back into the equation, we get:
Δx = v_ix * t
Δx = 5.00 m/s * 0.39 s
Δx ≈ 1.95 m
Therefore, the player must start their jump approximately 1.95 meters from the basket to reach their maximum height at the same time as reaching the basket.
To solve this problem, we can use the kinematic equations of motion. Let's break it down step by step.
(a) What vertical velocity does he need to rise 0.750 meters above the floor?
We can use the kinematic equation for vertical displacement to find the vertical velocity. The equation is:
y = v_iy * t + (1/2) * a * t^2
Here, y is the vertical displacement, v_iy is the initial vertical velocity, t is the time taken, and a is the acceleration in the y-direction. As the player jumps vertically, the acceleration in the y-direction will be due to gravity (approximately 9.8 m/s^2).
Given:
y = 0.750 m
a = -9.8 m/s^2 (negative because it is acting in the opposite direction of the jump)
Since the player starts with zero initial vertical velocity, the equation simplifies to:
0.750 m = (1/2) * -9.8 m/s^2 * t^2
Now we can solve for t:
t^2 = (2 * 0.750 m) / 9.8 m/s^2
t^2 = 0.153 m
t ≈ 0.391 s
So the time taken to rise 0.750 meters is approximately 0.391 seconds.
Now, to find the vertical velocity v_iy, we can use the equation:
v_fy = v_iy + a * t
Since the final vertical velocity (v_fy) is zero when he reaches his maximum height, we have:
0 m/s = v_iy - 9.8 m/s^2 * 0.391 s
Solving for v_iy:
v_iy = 9.8 m/s^2 * 0.391 s
v_iy ≈ 3.83 m/s
Therefore, the player needs a vertical velocity of approximately 3.83 m/s to rise 0.750 meters above the floor.
(b) How far from the basket (measured in the horizontal direction) must he start his jump to reach his maximum height at the same time as he reaches the basket?
Since the player maintains his horizontal velocity throughout the jump, we can use the equation:
Δx = v_ix * t
Here, Δx is the horizontal displacement, v_ix is the initial horizontal velocity, and t is the time taken.
Given:
v_ix = 5.00 m/s (horizontal velocity)
The time taken (t) is the same as the time taken to reach the maximum height, which we calculated in part (a) as approximately 0.391 seconds.
Now we can find the horizontal displacement:
Δx = 5.00 m/s * 0.391 s
Δx ≈ 1.96 m
Therefore, the player must start his jump 1.96 meters away from the basket (measured in the horizontal direction) in order to reach his maximum height at the same time as he reaches the basket.