A ball is thrown vertically upward from the ground with an initial velocity of 110ft/sec. Use h=−16t2+v0t+h0
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To solve this problem, we'll use the equation for vertical motion:
h = -16t^2 + v0t + h0
Where:
h: Height of the ball above the ground (in feet)
t: Time (in seconds)
v0: Initial velocity of the ball (in feet per second)
h0: Initial height of the ball (in feet)
In this case, the ball is thrown vertically upwards from the ground, so the initial height (h0) is 0.
Given information:
v0 = 110 ft/sec
h0 = 0 ft
Now, let's find out how long it takes for the ball to reach its highest point (when it starts falling back down). To find this, we'll set the final height (h) to 0 and solve for t:
0 = -16t^2 + 110t + 0
Since the equation is quadratic, we can solve it by factoring or using the quadratic formula. Let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
For this equation, a = -16, b = 110, and c = 0. Plugging these values into the quadratic formula:
t = (-110 ± √(110^2 - 4(-16)(0))) / (2(-16))
Simplifying:
t = (-110 ± √(12100)) / (-32)
t = (-110 ± 110) / (-32)
Now, we have two solutions for t. Let's calculate both:
t1 = (-110 + 110) / (-32) = 0
t2 = (-110 - 110) / (-32) ≈ 6.875
From these solutions, we can see that t1 = 0 and t2 = 6.875. Since the ball starts at the ground, t = 0 represents the initial position. Therefore, t2 = 6.875 is the time it takes for the ball to reach its highest point.
Now, let's find the maximum height (h) reached by the ball. We'll use the equation h = -16t^2 + v0t + h0, where t = 6.875:
h = -16(6.875)^2 + 110(6.875) + 0
h ≈ 1007.03 ft
Therefore, the maximum height reached by the ball is approximately 1007.03 feet.
Don't worry, I'm here to help you understand! To solve this problem, we can use the equation for the height of an object in free fall, which is given by h = -16t^2 + v0t + h0.
In this equation:
- h represents the height of the ball at any given time.
- t represents the time elapsed since the ball was thrown.
- v0 represents the initial velocity of the ball.
- h0 represents the initial height of the ball.
Here, we are given that the ball is thrown vertically upward from the ground, so the initial height (h0) is 0. The initial velocity (v0) is given as 110 ft/sec.
Now, let's break down the steps to find the answer:
Step 1: Determine the equation for the height (h) of the ball as a function of time (t).
Using the given values, the equation becomes:
h = -16t^2 + 110t
Step 2: Find the maximum height reached by the ball.
To find the maximum height, we know that the velocity of the ball is zero at its peak. So, we need to find the time (t) when the velocity (v) is zero.
Given that the initial velocity (v0) is 110 ft/sec, we can use the equation: v = v0 - 32t
Setting v = 0, we get:
0 = 110 - 32t
32t = 110
t = 110/32
t ≈ 3.44 seconds
Substitute t ≈ 3.44 seconds into the equation h = -16t^2 + 110t to find the maximum height (h).
Step 3: Calculate the maximum height (h).
Using the equation h = -16t^2 + 110t and substituting t ≈ 3.44 seconds:
h ≈ -16(3.44)^2 + 110(3.44)
h ≈ -16(11.8336) + 378.4
h ≈ -189.3376 + 378.4
h ≈ 189.0624 ft
So, the maximum height reached by the ball is approximately 189.0624 feet.
well, then h = -16t^2 + 110t
perhaps you might have a question in there somewhere?