just use your basic equations of motion:
h(t) = h0 + v0*t - 4.9t^2
v(t) = v0 - 9.8t
show what you get.
A blue ball is thrown upward with an initial speed of 23 m/s, from a height of 0.5 meters above the ground. 2.8 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 6.6 m/s from a height of 28.9 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s2.
1) What is the maximum height the blue ball reaches?
2) What is the height of the red ball 3.64 seconds after the blue ball is thrown?
3) How long after the blue ball is thrown are the two balls in the air at the same height?
4) Which statement is true about the blue ball after it has reached its maximum height and is falling back down?
a. The acceleration is positive and it is speeding up
b. The acceleration is negative and it is speeding up
c. The acceleration is positive and it is slowing down
d. The acceleration is negative and it is slowing down
Thank you!!!
2 answers
2. T = 3.64-2.8 = 0.84s.l
h = ho - (Vo*t + 0.5g*T^2).
h = 28.9 - (6.6*0.84 + 4.9*0.84^2),
4a. acceleration is positive and it is speeding up.
h = ho - (Vo*t + 0.5g*T^2).
h = 28.9 - (6.6*0.84 + 4.9*0.84^2),
4a. acceleration is positive and it is speeding up.