Three students have been studying relative motion and decide to do an experiment to demonstrate their knowledge. The experiment plan calls for Jane to drive her pickup in a straight line across the parking lot at a constant speed of 11.2 m/s. Fred is in the back of the truck and throws a baseball backward and upward at an angle θ out the back of the truck. Sue observes the flight of the ball while standing nearby in the parking lot.
(a) If Fred can throw the ball 31.8 m/s, at what angle relative to the horizontal should he throw the ball in order for Sue to see the ball travel vertically upward?
(b) If Fred throws the ball at this angle, how high does Sue observe it to travel above the level at which it was thrown?
a. He has to throw it forward...horizontal velocity=11.2
31.8cosTheta=11.2
solve for theta.
b. InitialKE=1/2 m (31.8SinTheta)2
Initial PE=0
final KE vertical=0
final PE=initial KE
mgh=1/2 m (31.8*SinTheta)^2
solve for h.
Thank you!
To solve this problem, we will break it down into two parts:
(a) To determine the angle at which Fred should throw the ball for Sue to see it travel vertically upward, we need to consider the horizontal component and vertical component of the ball's velocity.
Let's assume the angle Fred throws the ball relative to the horizontal is θ. The horizontal component of the ball's velocity is given by Vx = V * cos(θ), where V is the magnitude of the ball's velocity (31.8 m/s) and θ is the angle.
In this case, we want the vertical component of the velocity to be zero, which means Vy = 0. The vertical component of the ball's velocity is given by Vy = V * sin(θ).
Since we have Vx and Vy, we can set up two equations:
Vx = V * cos(θ)
Vy = V * sin(θ)
Given V = 31.8 m/s, we can solve for θ.
Vx = V * cos(θ)
31.8 m/s * cos(θ) = 11.2 m/s
Dividing both sides by 31.8 m/s, we get:
cos(θ) = 11.2 m/s / 31.8 m/s
cos(θ) = 0.351
Taking the inverse cosine (arc cosine) of both sides, we find:
θ = cos^(-1)(0.351)
θ ≈ 69.46 degrees
Therefore, Fred should throw the ball at an angle of approximately 69.46 degrees relative to the horizontal for Sue to see the ball travel vertically upward.
(b) To determine how high Sue observes the ball to travel above the level at which it was thrown, we can use the equation for the time of flight of a projectile:
T = 2 * Vy / g
where T is the total time of flight, Vy is the initial vertical component of the velocity, and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Since Vy = V * sin(θ), we can substitute it into the equation:
T = 2 * (V * sin(θ)) / g
Given V = 31.8 m/s and θ ≈ 69.46 degrees, we can calculate T.
T = 2 * (31.8 m/s * sin(69.46 degrees)) / 9.8 m/s^2
T ≈ 3.42 seconds
Now, we can calculate the maximum height (h) using the equation:
h = Vy * T - (1/2) * g * T^2
Given Vy = V * sin(θ), T ≈ 3.42 seconds, and g ≈ 9.8 m/s^2, we can calculate h.
h = (31.8 m/s * sin(69.46 degrees)) * 3.42 seconds - (0.5) * 9.8 m/s^2 * (3.42 seconds)^2
h ≈ 62.7 meters
Therefore, Sue observes the ball to reach a height of approximately 62.7 meters above the level at which it was thrown.
To solve this problem, we can break it down into two parts: determining the angle at which Fred should throw the ball for Sue to observe it traveling vertically upward, and then calculating the height Sue observes the ball to travel above the level at which it was thrown.
(a) To determine the angle at which Fred should throw the ball, we need to consider the relative motion between Fred's throw and Jane's truck. The ball's velocity is composed of two components: one in the horizontal direction due to the truck's motion, and one in the vertical direction due to Fred's throw.
To simplify this, we can consider the ball's motion with respect to an observer on the ground. Since we want Sue to see the ball travel vertically upward, the vertical component of the ball's velocity should be zero. We can use trigonometry to find the angle at which this occurs.
Let's assume the angle Fred throws the ball is given by θ relative to the horizontal. The vertical component of the ball's velocity is then given by v_y = v * sin(θ), where v is the magnitude of the ball's velocity (31.8 m/s) and θ is the angle relative to the horizontal.
Since we want Sue to see the ball travel vertically upward, the vertical component of the ball's velocity should be zero. Therefore, we set v_y = 0 and solve for θ:
0 = v * sin(θ)
sin(θ) = 0
Since sin(θ) = 0 only when θ = 0° or θ = 180°, we can conclude that to make the ball travel vertically upward relative to Sue, Fred should throw the ball at an angle of θ = 0° or θ = 180°.
(b) Now that we have determined the appropriate angle for the throw, we can calculate the height Sue observes the ball to travel above the level it was thrown.
Since the vertical component of the ball's velocity is zero, the time it takes for the ball to reach its highest point is given by the time it takes for the ball to stop moving upward and start moving downward. This can be calculated using the formula:
t = (2 * v_y) / g
where g is the acceleration due to gravity (approximately 9.8 m/s²).
Once we have the time, we can use it to calculate the maximum height reached by the ball using the formula:
h = v_y * t - (1/2) * g * t²
Substituting v_y = 0 and solving:
h = 0 - (1/2) * g * t²
h = - (1/2) * g * t²
Since the ball was thrown vertically upward, the height Sue observes the ball to travel above the level it was thrown is given by -(1/2) * g * t².
Note that the negative sign indicates that the ball is traveling downwards at its highest point. If we assume the level at which the ball was thrown is the same as Sue's observation point, then the height Sue observes the ball to travel is given by -(1/2) * g * t².
Please note that the answer will depend on the specific value of the magnitude of the ball's velocity (v) and the acceleration due to gravity (g) used in the calculations.