A T-ball player hits a ball from a tee that is 0.6m tall. The height of the ball at a given time is modelled by the function `h(t)=-4.9t^2+7t+0.6` .
where height, h(t) is in metres and time, t, is in seconds.
A) What will the height be after 1s?
b) When will the ball hit the ground?
1 Answer | Add Yours
`h(t)= -4.9t^2 + 7t +0.6`
`` a) we need to find the height of the ball after t= 1 second.
We will substitute with t= 1.
==> h(1)= -4.9(1) + 7(1)+ 0.6 = 2.7 m
Then the height after 1 second is 2.7 meters.
b)When will the ball hit the ground.
When the ball hit the ground then the height h= 0.
==> `h(t)= -4.9t^2 + 7t + 0.6 = 0 `
`==gt -4.9t^2 + 7t + 0.6 = 0`
Now we will solve for t.
==>`t1= (-7+sqrt(49+4*0.6*4.9))/(2*-4.9) = (-7+7.7949)/(-9.8)= -0.0811 ( Not valid) `
`==gt t2= (-7-7.7974)/(-9.8) = 1.9143` seconds.
Then the time for the ball to hit the ground is 1.9143 seconds.
Join to answer this question
Join a community of thousands of dedicated teachers and students.Join eNotes