a baseball is hit straight up from a height of 6 feet with an initial velocity of90 feet per second. given the equations=-16t^2 +v0t + s0, that models the height of the ball,s, as a function of time ,t where the initial height of the ball is s0 and the initial velocity is v0. How high is the ball after 4 seconds?
s=-16(4)^2 +90(4)+6
s=-256++366 +6
s=110
110 feet after 4 seconds
is that correct?
Thank you for checking my work
Yes, agree
thank you!
You are welcome :)
Yes, your calculation is correct. To find the height of the ball after 4 seconds, you can substitute the values into the equation:
s = -16t^2 + v0t + s0
Where:
t = 4 seconds
v0 = 90 feet per second (initial velocity)
s0 = 6 feet (initial height)
Therefore,
s = -16(4)^2 + 90(4) + 6
s = -256 + 360 + 6
s = 110
So, the ball is at a height of 110 feet after 4 seconds. Well done!