This is really urgent so please please please help.
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
Is there a difference but, anyway I didn't make this clear on the last post.
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
H(t)=-4.9t^2+30t+53
H(t)=???
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.
Please Help and Thank You very much =)
acceleration downward due to gravity is 10m/s to be exact 9.8m/s
this is enough to get you to start thinking
h(t) = -4.9t^2 + 30t + 53
a) ball goes up, comes back down to the top of the tower. So, we want
53 = -4.9t^2 + 30t + 53
0 = -4.9t^2 + 30t
0 = t(-4.9t + 30)
so, t=0 (at the start) or t = 6.12 (as it comes back down)
If t(-4.9t+30)=0, either
t=0
or
-4.9t+30 = 0
That is, t = 6.12
If you can't solve a factored expression, you have some review to do.
b) same calculation but, t is between 0 and 6.12. That is 0 < t < 6.12
c) same calculation, but restricting t to positive values, t>6.12
Naturally, we could also restrict t to the point where height >= 0.
d) vertex of any parabola is where x = -b/2a = -30/-9.8 = 3.06
you know from the quadratic formula that x = -b/2a ± sqrt(blah blah)
Parabolas are symmetric, so the vertex is midway between the roots, which are equally spaced around x = -b/2a
h(3.06) = 98.9
To solve these problems, we need to substitute the given values into the equation for the height:
H(t) = -4.9t^2 + vt + h
where v is the initial velocity of the ball (30 m/s) and h is the initial height of the ball (53 m).
a) To find the time at which the ball's height equals the height of the tower, we set H(t) equal to the height of the tower (53 m) and solve for t:
-4.9t^2 + 30t + 53 = 53
Simplifying the equation, we get:
-4.9t^2 + 30t = 0
Factoring out t, we have:
t(-4.9t + 30) = 0
So either t = 0 or -4.9t + 30 = 0. Solve the second equation for t:
-4.9t + 30 = 0
-4.9t = -30
t = -30 / -4.9
t ≈ 6.12 seconds
Therefore, the time at which the ball's height equals the height of the tower is approximately 6.12 seconds.
b) To find the time at which the ball's height is greater than the height of the tower, we need to find when H(t) is greater than 53 m. This means we need to solve the inequality:
-4.9t^2 + 30t + 53 > 53
Simplifying the inequality, we get:
-4.9t^2 + 30t > 0
Factoring out t, we have:
t(-4.9t + 30) > 0
So either t > 0 and -4.9t + 30 > 0, or t < 0 and -4.9t + 30 < 0. Solve each inequality separately:
For t > 0 and -4.9t + 30 > 0:
-4.9t + 30 > 0
-4.9t > -30
t < 30 / 4.9
t < 6.12 seconds
For t < 0 and -4.9t + 30 < 0:
-4.9t + 30 < 0
-4.9t < -30
t > 30 / 4.9
t > 6.12 seconds
Therefore, the ball's height is greater than the height of the tower for t < 0 and t > 6.12 seconds (excluding t = 6.12 seconds).
c) To find the time at which the ball's height is less than the height of the tower, we need to find when H(t) is less than 53 m. This means we need to solve the inequality:
-4.9t^2 + 30t + 53 < 53
Simplifying the inequality, we get:
-4.9t^2 + 30t < 0
Factoring out t, we have:
t(-4.9t + 30) < 0
So either t > 0 and -4.9t + 30 < 0, or t < 0 and -4.9t + 30 > 0. Solve each inequality separately:
For t > 0 and -4.9t + 30 < 0:
-4.9t + 30 < 0
-4.9t < -30
t > 30 / 4.9
t > 6.12 seconds
For t < 0 and -4.9t + 30 > 0:
-4.9t + 30 > 0
-4.9t > -30
t < 30 / 4.9
t < 6.12 seconds
Therefore, the ball's height is always greater than the height of the tower.
d) To find the time at which the ball reaches its maximum height, we need to find the vertex of the quadratic equation. The vertex can be found using the formula:
t = -b / (2a)
where a = -4.9 and b = 30. Substituting these values into the formula:
t = -30 / (2 * -4.9)
t = 30 / 9.8
t ≈ 3.06 seconds
Therefore, the ball reaches its maximum height at approximately 3.06 seconds.