A science student is riding on a flatcar of a train traveling along a straight, horizontal track at a constant speed of 12.8 m/s. The student throws a ball into the air along a path that he judges to make an initial angle of 67.0° with the horizontal and to be in line with the track. The student's professor, who is standing on the ground nearby, observes the ball to rise vertically. How high does she see the ball rise?
v₀=v•sinα
h=v₀²/2g= (v•sinα)²/2g= (12.8•sin67º)²/2•9.8=7.08 m.
To find out how high the professor sees the ball rise, we need to analyze the motion of the ball both horizontally and vertically. Let's break it down step by step:
1. Find the initial vertical velocity (Vy0) of the ball:
The student throws the ball vertically, so there is no initial horizontal velocity (Vx0 = 0). We only need to calculate the initial vertical velocity.
Vy0 = V0 * sin(θ)
Vy0 = 12.8 m/s * sin(67.0°)
2. Find the time it takes for the ball to reach its maximum height:
At the maximum height, the vertical velocity becomes zero (Vy = 0). We can use the equation:
Vy = Vy0 + (-g) * t
0 = Vy0 - 9.8 m/s^2 * t
Solve for t to find the time of flight.
3. Calculate the time it takes for the ball to reach its maximum height:
t = Vy0 / g
Plug in the value of Vy0 calculated from step 1 and g = 9.8 m/s^2.
4. Find the height the ball reaches:
The height the ball reaches can be calculated using the formula:
H = H0 + Vy0 * t + (1/2) * (-g) * t^2
H0 = 0 since we are measuring the height from the ground level.
Plug in the values of Vy0 and t calculated from the previous steps to find the maximum height the professor observes.
By following these steps, you should be able to find the height the professor sees the ball rise.