David drops a ball from a bridge at an initial height of 100 meters. 1. What is the height of the ball to the nearest tenth of a meter exactly 3 seconds after he releases the ball? 2. How many seconds after the ball is released will it hit the ground? I have 3 questions, and I answered the first two just fine, but this kind of stuff my brain just can't get. So if you could explain it with your answer, that would be amazing. This has to be solved using h(t)= 1/2gt^2+v0t+h0I know that the v0 is 0, because there's no force given, and that g is -9.8 m/s^2 because it's meters being used, and I'm pretty sure h0 is 100, I just don't know how to put it all together, or where the 3 seconds after he releases the ball is supposed to go. Thanks for any help, and whoever answers correctly, I have either 10 or 15 points for you! (I think I might have pressed 15 but it could have also been 10 and it won't let me go back and look)

[tex] h(t) = \frac{1}{2}gt^{2} + v0t + h0 [/tex]

this equation is the given equation that will determine the high of an object using the time. This equation comes from physic though, so if you have a chance to study physic classes, you would have more understanding of this formula
g represents the gravity, -9.8 m/s^2 is the acceleration.
t represents the second. 
you just let t = 3, plug them into the equation, v0: initial speed is 0 because the ball was dropping, not throwing down. 
[tex] h(3) = \frac{1}{2}(-9.8)(3)^{2} + (0)(3) + 100 [/tex]
[tex] h(3) = 55.9 m [/tex] is the height of the ball at 3 second. 
The second question asks you how many seconds will the ball hit the ground. When the ball touches the ground, that means the height is 0, but you don't how long was the dropping down.

Still, we use the equation [tex] h(t) = \frac{1}{2}gt^{2} + v0t + h0 [/tex]
Note: v0 is always 0 in this situation as I mentioned above. This equation does not involve the speed of ball during the dropping time, so no need to worry about it

[tex] 0 = \frac{1}{2}(-9.8)t^{2} + (0)t + 100 [/tex]
Now we solve this equation to find t.
[tex] 0 = -4.9t^{2} + 100 [/tex]
[tex]  -4.9t^{2} = - 100 [/tex]
[tex]\sqrt{t^{2}} = \sqrt{ \frac{-100}{-4.9} } ==\ \textgreater \ t = 4.5s [/tex] is the total time for the ball to hit the ground