A more accurate answer for the required velocity is sqrt(2gH) = 31.3 m/s
The time in the air is 2*31.3/9 = 6.39 s
y = 31.3 t -4.9 t^2 is height vs. time.
It is an upside down parabola with a maximum at t = 3.2 s
v = 31.3 - 9.8 t is velocity vs time
It is a straight line that goes through zero at t = 3.2 s
a = -9.8 is acceleration vs time
We can't draw graphs for you here.
i have this problem, it asks:
(a) With what speed must a ball be thrown vertically from ground level to rise to a maximum height of 50m?
answer: 31m/s
(b)How long will it be in the air?\
answer: 6.4s
(c) Sketch graphs of y, v, and a vs. t for the ball. On the first two graphs, indicate the time at which 50m is reached.
i already solved for a and b but i cannot figure how the graph for c would look.
any help is appreciated!
1 answer
2 answers