A ball is shot from the ground into the air. At a height of 9.1 m, the velocity is observed to be v = (6.6 + 5.7 ) m/s, with horizontal and upward.
(a) To what maximum height does the ball rise?
m
(b) What total horizontal distance does the ball travel?
m
(c) What is the magnitude of the ball's velocity just before it hits the gound?
m/s
(d)What is the direction of the ball's velocity just before it hits the ground?
° relative to the horizontal
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Yall wrong
To solve this problem, we can use the equations of motion for projectile motion. Let's break down each part of the problem:
(a) To find the maximum height reached by the ball, we need to determine the vertical component of its initial velocity. We are given that the observed velocity at a height of 9.1 m is v = (6.6 + 5.7) m/s upward. This means the upward velocity component is 5.7 m/s. At the maximum height, the ball's vertical velocity component will be zero since it momentarily comes to a stop.
To find the time it takes for the ball to reach its maximum height, we can use the equation:
v = u + at
Where:
v = final vertical velocity (0 m/s)
u = initial vertical velocity (5.7 m/s)
a = acceleration due to gravity (-9.8 m/s^2)
t = time
Rearranging the equation to solve for time, we have:
t = (v - u) / a
Substituting the given values, we get:
t = (0 - 5.7) / -9.8
Solving for time, t = 0.582 seconds.
Now, we can calculate the maximum height using the equation:
h = u * t + (1/2) * a * t^2
Where:
h = maximum height
u = initial vertical velocity (5.7 m/s)
t = time (0.582 seconds)
a = acceleration due to gravity (-9.8 m/s^2)
Substituting the values, we have:
h = 5.7 * 0.582 + (1/2) * (-9.8) * (0.582)^2
Simplifying the equation, we find that the maximum height is approximately 2.52 meters.
Therefore, the maximum height reached by the ball is 2.52 meters.
(b) To find the total horizontal distance traveled by the ball, we can use the equation:
d = v * t
Where:
d = horizontal distance
v = horizontal velocity component of the ball (6.6 m/s)
t = time (0.582 seconds)
Substituting the values, we have:
d = 6.6 * 0.582
Simplifying the equation, we find that the horizontal distance traveled by the ball is approximately 3.84 meters.
Therefore, the total horizontal distance covered by the ball is 3.84 meters.
(c) To find the magnitude of the ball's velocity just before it hits the ground, we need to calculate the resultant velocity vector.
The resultant velocity vector can be calculated using the Pythagorean theorem:
resultant velocity = sqrt((horizontal velocity)^2 + (vertical velocity)^2)
Using the given values, we have:
resultant velocity = sqrt((6.6)^2 + (5.7)^2)
Calculating the value, we find that the resultant velocity is approximately 8.61 m/s.
Therefore, the magnitude of the ball's velocity just before hitting the ground is 8.61 m/s.
(d) To find the direction of the ball's velocity just before it hits the ground relative to the horizontal, we can use trigonometry.
The angle, θ, can be calculated using the equation:
θ = arctan(vertical velocity / horizontal velocity)
Using the given values, we have:
θ = arctan(5.7 / 6.6)
Calculating the value, we find that the angle is approximately 40.81 degrees.
Therefore, the direction of the ball's velocity just before it hits the ground is approximately 40.81 degrees relative to the horizontal.
6.7²/(2g) + 9.1 = 11.39 m
b) t = 2√[2H/g] = 3.05 sec; X = 6.6*t = 20.125 m
c) Vf = (6.6i - 6.7j) m/s
d) Θ = arctan[j/i] = arctan[-6.7/6.6] = -45.43°