a ball is thrown straight up at an initial velocity of 54 feet per second, the height of the ball at T seconds after it is thrown is given by the formula h(t)=5t-12t^2, how many seconds after the ball is thrown will it return to the ground
+33661 Hmm! If u is the initial upward velocity and g is the gravitational acceleration, then
\(\frac{dh}{dt}= u-g\times t\)
This means that
\(h(t)=u\times t-\frac{g}{2}\times t^2\)
So, either u is 54 or the initial velocity is 5 ft/sec. Which is it?
(Also, it looks like g = 24 ft/sec^2 here.)
+1316 I did this and got 4.5f/1s^2
I also find out the ball goes to 60 feet high before it start coming back
If Alan not say the gravity is 24f/s^2 I probably not notice.
So know I wonder why this is not 32f/s^2. I think maybe the ball have a low terminal velocity. To find out that we have to know the drag coefficient and the mass of the ball. I not really know how to do it for this.
Alan can a drag coefficient make it seem like the gravity is weaker?
+1316 I not answer this right. It take 4.5 seconds to hit the ground is what I mean.
