A science student is riding on a flatcar of a train along a strait, horizontal track at a constant speed of 12.0 m/s. The student throws a ball into the air along a path he judges to make an initial angle 65. deg 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 he see the ball rise?

Question

A science student is riding on a flatcar of a train along a strait, horizontal track at a constant speed of 12.0 m/s. The student throws a ball into the air along a path he judges to make an initial angle 65. deg 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 he see the ball rise?

Answer 1

This reduces to finding the max height of a ball thrown upward with a speed of

12.0 sin65° m/s

I'm sure you've done that kind before.

Answer 2

To calculate how high the professor sees the ball rise, we can break the initial velocity of the ball into horizontal and vertical components.

First, let's calculate the initial vertical velocity of the ball. We know that the ball is thrown vertically upwards, so its initial vertical velocity (Viy) is the same as its final vertical velocity just before it reaches its highest point. At the highest point, the vertical velocity becomes zero since the ball momentarily stops before falling back down.

The initial vertical velocity (Viy) can be calculated using the given launch angle (θ) and the initial velocity of the ball (Vi). The formula to calculate the initial vertical velocity is:

Viy = Vi * sin(θ)

Next, we need to find the time it takes for the ball to reach its highest point. Since the initial vertical velocity (Viy) is known, we can use the following equation to calculate the time (t) it takes for the ball to reach its highest point:

Vf = Viy + g * t

At the highest point, the final vertical velocity (Vf) becomes zero. Thus, the equation becomes:

0 = Viy + g * t

Now, we can isolate t in this equation:

t = -Viy / g

Since acceleration due to gravity (g) is approximately 9.8 m/s², we can plug in the values to calculate the time it takes for the ball to reach its highest point.

Next, we can use the time (t) to calculate the height (h) the ball reaches. The formula for calculating the height at a certain time using the initial vertical velocity and time is:

h = Viy * t + (1/2) * g * t²

However, in this case, we are interested in the maximum height reached, so we need to divide t by 2 to get the time at the highest point:

t_highest = t / 2

Finally, we can substitute the calculated values into the equation to find the height (h) at the highest point.

Please provide the value of the initial velocity of the ball (Vi) to continue.