Yes to your question ab9ut the rise and fall time.
Time to rise or fall = 2.2s/2 = 1.1 s
Time to fall T is also iven by
g T = Vo, so
T = Vo/g = 1.1 = Vo/9.8
Vo = 9.8 m/s^2 * 1.1 s = 10.8 m/s
Distance travelled up = Vav*T = Vo/(2T)
= 5.4 m/s*1.1 s = 5.4 m
You throw a beambag in the air and catch it 2.2 s later.
GIVEN:
t = 2.2 s
vi = 0 m/s
a. How high did it go?
d = vit + 1/2 at^2
d = 1/2 (9.81)(2.2)^2
d = 23.7402
b. What was its initial velocity?
I don't know if im right but to calculate the velocity, can you multiply 9.81 x 1.1s = 10.8 m/s.
3 answers
The initial velocity is 0 because it is starting in your hand, not moving
initial velocity is 11m/s
1 answer