Asked by Emily
A baseball is thrown upwards from a height of 2 meters with an initial velocity of 10 meters per second. Use the fact that -9.8 meters per second squared is constant acceleration due to gravity to answer the following:
The maximum height of the ball
The time it takes for the ball to hi the ground.
The maximum height of the ball
The time it takes for the ball to hi the ground.
Answers
Answered by
Reiny
height = -4.9t^2 + 10t + 2
d(height)/dt = -9.8t + 10
= 0 at max height
9.8t = 10
t = 10/9.8
height = -4.9(10/9.8)^2 + 10(10/9.8) + 2 = appr 7.102 m is the max height
when it reaches the ground
0 = -4.9t^2 + 10t + 2
4.9t^2 - 10t- 2 = 0
t = (10 ± √139.2)/9.8 = 1.306 seconds or a negative
check my arithmetic
d(height)/dt = -9.8t + 10
= 0 at max height
9.8t = 10
t = 10/9.8
height = -4.9(10/9.8)^2 + 10(10/9.8) + 2 = appr 7.102 m is the max height
when it reaches the ground
0 = -4.9t^2 + 10t + 2
4.9t^2 - 10t- 2 = 0
t = (10 ± √139.2)/9.8 = 1.306 seconds or a negative
check my arithmetic
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