V = Vo + g*Tr = 0,
14.72 + (-9.8)Tr = 0,
Tr = 1.50 s. = Rise time or time to reach max ht.
a. Disp. = 0.5g*Tf^2 = 4.9(2-1.5)^2 = 1.225 m.
Tf = Fall time.
b. V = Vo + g*t = 0 + 9.8(2-1.5) = 4.9 m/s
c. Since the ball starts falling after 1.5 seconds in air, the acceleration after 2 s = g = 9.8 m/s^2.
a ball is thrown up into the air with an initial velocity of 14.72 m/s What is the displacement, velocity, and acceleration after 2 seconds?
2 answers
Correction:
a. V^2 = Vo^2 + 2g*h = 0,
14.72^2 + (-19.6)h = 0,
h = 11.1 m. = max ht.
Disp. = ho - 0.5g*Tf^2 = 11.1 - 4.9*(2-1.5)^2 = 9.875 m.
a. V^2 = Vo^2 + 2g*h = 0,
14.72^2 + (-19.6)h = 0,
h = 11.1 m. = max ht.
Disp. = ho - 0.5g*Tf^2 = 11.1 - 4.9*(2-1.5)^2 = 9.875 m.
1 answer