Final answer:
The ball must be thrown with an initial speed of approximately 31.3 m/s to achieve a maximum height of 50 meters, and it will be in the air for approximately 6.38 seconds. The y vs. t graph is parabolic, the v vs. t graph is a straight line with negative slope, and the a vs. t graph is a horizontal line.
Step-by-step explanation:
The student wants to know with what speed a ball must be thrown vertically from ground level to rise to a maximum height of 50m, how long it will be in the air, and wishes to see graphs of position (y), velocity (v), and acceleration (a) versus time (t) for the ball's motion.
To find the speed (v) necessary to reach a height (h) of 50m, one can use the equation of motion for uniform acceleration:
h = (v^2) / (2g)
where, g is the acceleration due to gravity (9.81 m/s^2). Rearranging and solving for v yields initial speed v:
v = sqrt(2gh) = sqrt(2 * 9.81 * 50) = 31.3 m/s (approx)
To find the time (t) the ball is in the air, we consider that the time to rise to the peak is half the total time in the air.
t = (2 * v) / g = (2 * 31.3) / 9.81 = 6.38 seconds (approx for round trip)
The sketch of the graphs would show: (a) a parabolic trajectory for y, with y increasing to 50m and then decreasing back to 0, (b) a linear decrease for velocity v, from 31.3 m/s to -31.3 m/s (assuming down is negative), and (c) a constant negative value for acceleration a, representing the acceleration due to gravity.