Projectile motion question: how long is ball in the air?

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sirkal

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You are playing soccer with some friends while some children are plying hide-and-seek nearby. The seeker is busy counting at a rate of one whole number per second. When you hear the seeker say the number “4,” you kick the ball high in the air. When the seeker says “7,” the ball is 20 feet in the air. How many seconds will the ball be in the air in total before it hits the ground again?
 
sirkal said:
You are playing soccer with some friends while some children are plying hide-and-seek nearby. The seeker is busy counting at a rate of one whole number per second. When you hear the seeker say the number “4,” you kick the ball high in the air. When the seeker says “7,” the ball is 20 feet in the air. How many seconds will the ball be in the air in total before it hits the ground again?
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sirkal said:
You are playing soccer with some friends while some children are plying hide-and-seek nearby. The seeker is busy counting at a rate of one whole number per second. When you hear the seeker say the number “4,” you kick the ball high in the air. When the seeker says “7,” the ball is 20 feet in the air. How many seconds will the ball be in the air in total before it hits the ground again?
Click to expand...
What have you learned about projectile motion? Put the point t=3, h=20 into an equation for which at t=0, h=0 [or alternatively, use t=7 and t=4], and solve for whatever you can. Then show us what you have done, and what other formulas you know.
 
I used the equation S = 1/2gt2, with t = 3 and s = 20. I then used the equation -u = u - gt and arrived at the answer 8.667 minutes. Can someone confirm this for me?
 
I get a different answer (and it would be in seconds, not minutes).

Please show the details of what you did. Since the formula you mention applies only to objects that fall from rest, not to projectiles with a given initial velocity, your work is probably wrong.

Can you list all the formulas you have learned, so we can be sure what we can help you to use? I solved the problem an entirely different way that you probably couldn't use.
 
The formula S = 1/2gt2 was wrong. The formula I used is S = u - 1/2gt2 with t = 3 and s = 20. I then used the equation -u = u - gt. That's how I arrived at 8.667 seconds.
 
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