The height(H) of an object that has been dropped or thrown in the air is given by:

H(t)=-4.9t^2+vt+h
t=time in seconds(s)
v=initial velocity in meters per second (m/s)
h=initial height in meters(m)

H=height
h=initial height
I didn't make this clear on the last post sorry.

A ball is thrown vertically upwardd from the top of the Leaning Tower of Pisa (height=53m) with an initial velocity of 30m/s. Find the time(s) at which:
a) the ball's height equals the hight of the tower
b) the ball's height is greater than the height of the tower
c) the ball's height is less than the height of the tower
d)the ball reaches its maximum height

I don't know how to do this problem.

a)H(t)=-4.9t^2+30t+53
..H(t)=???

Please Help and Thank You very much =)

To find the time(s) at which the ball's height equals the height of the tower, you need to set the height equation equal to the height of the tower and solve for t.

So for part a), you have H(t) = -4.9t^2 + 30t + 53 and you want to find the time when the ball's height equals the height of the tower (53m). Therefore, you can set H(t) equal to h and solve for t:

-4.9t^2 + 30t + 53 = 53

To solve this quadratic equation, first bring everything to one side of the equation to get:

-4.9t^2 + 30t = 0

Next, factor out t from the equation:

t(-4.9t + 30) = 0

Now, you have two possibilities for t:

1) t = 0
2) -4.9t + 30 = 0

The first solution, t = 0, represents the initial time when the ball was thrown.

To find the second solution, solve -4.9t + 30 = 0 for t:

-4.9t = -30
t = -30 / -4.9 ≈ 6.12 seconds

Therefore, the ball's height equals the height of the tower at t = 0 seconds (initial time) and t ≈ 6.12 seconds.

Now let's move on to part b) and find the time(s) when the ball's height is greater than the height of the tower.

Since the height of the tower is 53m, we need to solve the inequality H(t) > 53. Using the same height equation H(t) = -4.9t^2 + 30t + 53, we have:

-4.9t^2 + 30t + 53 > 53

Simplifying this inequality:

-4.9t^2 + 30t > 0

Now, factor out t from the equation:

t(-4.9t + 30) > 0

To solve this inequality, we need to consider the signs of both t and -4.9t + 30.

Since t multiplied by any number greater than 0 is always positive, we know that t > 0 is one possible solution.

To find the other solution, solve -4.9t + 30 > 0:

-4.9t > -30
t > -30 / -4.9 ≈ 6.12 seconds

Therefore, the ball's height is greater than the height of the tower for t > 6.12 seconds.

Finally, for part c) - finding the time(s) when the ball's height is less than the height of the tower.

Using the same height equation H(t) = -4.9t^2 + 30t + 53, we need to solve the inequality H(t) < 53. Again, simplify the inequality:

-4.9t^2 + 30t + 53 < 53

-4.9t^2 + 30t < 0

Factor out t:

t(-4.9t + 30) < 0

Now, considering the signs of t and -4.9t + 30:

Since t multiplied by any number less than 0 is always negative, we know that t < 0 is one possible solution.

To find the other solution, solve -4.9t + 30 < 0:

-4.9t < -30
t < -30 / -4.9 ≈ -6.12 seconds

Therefore, the ball's height is less than the height of the tower for t < -6.12 seconds.

Finally, for part d) - finding the time when the ball reaches its maximum height.

The maximum height occurs at the vertex of the parabolic trajectory, which can be determined by finding the t-value at the axis of symmetry of the quadratic equation. The axis of symmetry formula is given by t = -b / (2a), where a = -4.9 and b = 30 (from -4.9t^2 + 30t + 53).

t = -30 / (2 * (-4.9))
t ≈ -3.06 seconds

Therefore, the ball reaches its maximum height at approximately t ≈ -3.06 seconds.

Please note that in this context, the negative value for time represents the time before the ball was thrown. The actual time it takes for the ball to reach its maximum height is approximately t ≈ 3.06 seconds after the ball is thrown upwards.

a. -4.9t^2+30t+53 = 53.

Use Quad. Formula and get t=6.12245 s.

b. -4.9t^2+30t+53 > 53.
Solve for t.

c. -4.9t^2+30t+53 < 53.
solve for t.

d. t = (V-Vo)/g.
t = (0-30) / -9.8 = 3.06 s.