A ball is thrown upward with an initial velocity of 96 feet per second from the top of a 100 foot building. (a) When well the ball attain its maximum height? (b) When will the ball hit the ground? Give a decimal answers correct to 2 decimal places.
the height relative to the ground will be
s(t) = 16t^2 + 90t + 100
a)
s'(t) = -32t + 90
= 0 for max of s(t)
32t = 90
t = 90/32 = 2.8125 seconds
b)
ball hits the ground when s(t) = 0
-16t^2 + 90t + 100=0
8t^2 - 45t - 50 = 0
solve for t using the quadratic formula , refect the negative result
Shouldn't 90t be 96t since 96 feet/second is the velocity?
To solve this problem, we can use the equation of motion for an object in free fall:
h(t) = h₀ + v₀t - 16t²
where:
h(t) is the height of the ball at time t
h₀ is the initial height (in this case, 100 ft)
v₀ is the initial velocity (in this case, 96 ft/s)
t is the time.
(a) To find the maximum height, we need to find the time at which the ball reaches its peak. At the maximum height, the velocity of the ball will be zero (v = 0).
Using the equation v(t) = v₀ - 32t, we can set v(t) = 0 and solve for t:
0 = v₀ - 32t
32t = v₀
t = v₀ / 32
Let's substitute the given values:
t = 96 ft/s / 32 ft/s²
t = 3 seconds
So, the ball will reach its maximum height after 3 seconds.
(b) To find when the ball will hit the ground, we can set the height h(t) to zero and solve for t:
h(t) = h₀ + v₀t - 16t² = 0
Using the quadratic formula, we have:
t = (-v₀ ± √(v₀² - 4(-16)(h₀))) / (2(-16))
Substituting the given values:
t = (-96 ± √(96² - 4(-16)(100))) / (2(-16))
Calculating the square root:
t = (-96 ± √(9216 + 6400)) / (-32)
t = (-96 ± √15616) / (-32)
t = (-96 ± 124.9) / (-32)
Dividing each term by -32:
t₁ = (96 - 124.9) / 32 ≈ -0.89 seconds (extraneous, time can't be negative)
t₂ = (96 + 124.9) / 32 ≈ 6.16 seconds
So, the ball will hit the ground after approximately 6.16 seconds.
In summary:
(a) The ball will attain its maximum height after 3 seconds.
(b) The ball will hit the ground after approximately 6.16 seconds.
To solve this problem, we can use the equations of motion of a free-falling object. The height and velocity of the ball can be determined at any given time using these equations.
(a) To find the time it takes for the ball to reach its maximum height, we need to determine when the ball's velocity becomes zero. At maximum height, the ball will momentarily stop moving upward before it starts to fall back down.
The equation to determine the height of a falling object as a function of time, neglecting air resistance, is given by:
h(t) = h0 + v0t - (1/2)gt^2
In this equation:
- h(t) represents the height of the object at time t
- h0 is the initial height of the object
- v0 is the initial velocity of the object
- g is the acceleration due to gravity, which is approximately 32.2 ft/s^2
Using the given values:
h0 = 100 ft (initial height)
v0 = 96 ft/s (initial velocity)
g = 32.2 ft/s^2 (acceleration due to gravity)
Let's solve for t when the velocity (v) is zero:
0 = v0 - gt
Rearranging the equation:
t = v0 / g
Substituting the known values:
t = 96 ft/s / 32.2 ft/s^2
Calculating:
t ≈ 2.98 seconds
Therefore, the ball will attain its maximum height approximately 2.98 seconds after being thrown.
(b) To find when the ball will hit the ground, we need to determine the time it takes for the ball to reach a height of 0 feet.
Using the equation: h(t) = h0 + v0t - (1/2)gt^2
Setting h(t) to 0, h0 to 100 ft, and solving for t:
0 = 100 ft + 96 ft/s * t - (1/2) * 32.2 ft/s^2 * t^2
Rearranging and simplifying the equation:
16.1t^2 - 96t - 100 = 0
We can solve this quadratic equation to find the two possible values of t. Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
For this equation:
a = 16.1
b = -96
c = -100
Calculating:
t ≈ 5.88 seconds (approximately)
So, the ball will hit the ground approximately 5.88 seconds after being thrown.