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.
initial velocity is 11m/s
To calculate the height the beambag reached and its initial velocity, we can use the basic equations of motion.
a. To find the height the beambag reached (d), we can use the equation:
d = vit + 1/2 at^2
Where:
d is the displacement (height) of the beambag
vi is the initial velocity of the beambag (which is 0 in this case)
a is the acceleration due to gravity (approximately 9.81 m/s^2)
t is the time it takes for the beambag to reach its maximum height (2.2 seconds in this case)
Plugging in the values into the equation, we get:
d = 0(2.2) + 1/2 (9.81)(2.2)^2
d = 0 + 1/2 (9.81)(4.84)
d = 1/2 (47.90)
d = 23.95 meters (rounded to two decimal places)
Therefore, the beambag reached a height of approximately 23.95 meters.
b. To find the initial velocity (vi), we can use the equation:
vi = vf - at
Where:
vi is the initial velocity of the beambag (what we're trying to find)
vf is the final velocity of the beambag at the top of its trajectory (which is 0 as it reaches the top)
a is the acceleration due to gravity (approximately 9.81 m/s^2)
t is the time it takes for the beambag to reach its maximum height (2.2 seconds in this case)
Plugging in the values into the equation, we get:
vi = 0 - (9.81)(2.2)
vi = -21.582 meters per second
Since the initial velocity of an object cannot be negative, we disregard the negative sign, and thus the initial velocity of the beambag is approximately 21.582 meters per second (rounded to three decimal places).
It's important to note that in the calculation for the initial velocity, you multiplied 9.81 by 1.1 seconds. However, the correct formula to use is vf = vi + at, where vf is the final velocity. Since we know the final velocity is 0 at the top of the motion, we can solve for the initial velocity.
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