A ball thrown upward reaches a height of 13m. How much time did it take to get to this height? How much time will take to fall to the ground?
consider energy.
InitialKineticEnergy=finalPotentialEnergy
1/2 mv^2=mg*13
V =sqrt(2*9.8*13) that is the initial velocity.
the average velocity going up is half that, so time up= 12/(1/2 sqrt(2*9.8*13)
grab your calculator.
Time down is the same as time up.
yhtrh
To find the time it took for the ball to reach a height of 13m, we can use the equation of motion for free-falling objects:
h = (1/2) * g * t^2
where:
h is the height
g is the acceleration due to gravity (approximately 9.8 m/s^2)
t is the time
Rearranging the equation to solve for time:
t = sqrt(2h / g)
Plugging in the values:
t = sqrt(2 * 13 / 9.8)
t = sqrt(26 / 9.8)
t ≈ sqrt(2.653)
t ≈ 1.63 seconds
Therefore, it took approximately 1.63 seconds for the ball to reach a height of 13m.
Now, to find the time it will take for the ball to fall to the ground, we can use the same equation. However, this time we will consider the initial height to be 0m.
t = sqrt(2 * 0 / 9.8)
t = sqrt(0 / 9.8)
t = 0 seconds
Since the initial height is 0m, the ball is already on the ground. Therefore, it will take 0 seconds for the ball to fall to the ground.