Question

David drops a ball from a bridge at an initial height of 100 meters.

Answer 1

$h(t) = \frac{1}{2}gt^{2} + v0t + h0$

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. 
$h(3) = \frac{1}{2}(-9.8)(3)^{2} + (0)(3) + 100$
$h(3) = 55.9 m$ 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 $h(t) = \frac{1}{2}gt^{2} + v0t + h0$
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

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