how do i find the height of a baseball if it is hit 270ft it takes the ball 4 secs to hit the ground. the velocity is 67.5 ft per sec, the acceleration is 16.88 feet per sec^2. the batter hits the ball 3.5ft off the ground. what is the height of the ball at its highest peek before it starts to drop?
formula for height
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tutor_joel
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I think you need to be a bit more specific with this problem. How is it EXACTLY stated? Because you're giving me a velocity and an acceleration. Which is fine, but there seems to be something missing here. Is there a launch angle? Are you using equations of motion or just straight math?
here is a formula that might help you.
distance= ((Initial velocity+Final velocity)(time))/2
In addition to all of the other stuff that you have, you also know that the acceleration due to gravity is -32.174 ft/s2, the final velocity is 0 ft/sec since it is going to the peak, and you know that the time is cut in half since you only need to know how far the baseball is going up.
distance= ((Initial velocity+Final velocity)(time))/2
In addition to all of the other stuff that you have, you also know that the acceleration due to gravity is -32.174 ft/s2, the final velocity is 0 ft/sec since it is going to the peak, and you know that the time is cut in half since you only need to know how far the baseball is going up.
tutor_joel
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Where is the acceleration from? The acceleration after contact with the bat? If that's the case, where is the velocity from? The final velocity? There is plenty of info to work this problem, but I need more specifics.
The 270 ft should be the total distance x traveled. (under the parabola)
The 270 ft should be the total distance x traveled. (under the parabola)
tutor_joel
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mathgeek said:
That is not true since he strikes the ball 3.5 ft off of the ground
LivingMath
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The horizontal distance the ball traveled is not important for this problem, since we're only interested in the vertical position of the ball.
The formula for the position of an object in projectile motion (eg. under constant acceleration from gravity) is
\(\displaystyle x_t = x_0 + tv_0 + \frac{1}{2}t^2a\)
In this case, we have the following information:
\(\displaystyle x_0 = 3.5\)
\(\displaystyle x_4 = 0\)
\(\displaystyle a = g = -32\) (rounded off for ease of calculation)
By evaluating the position equation at t = 4, you can find \(\displaystyle v_0\). Remember that this is the vertical velocity, not the horizontal velocity.
With \(\displaystyle v_0\), you have enough information to make a parabolic equation, which you can use to get the maximum height in several ways. Please let us know if you need help from here.
The formula for the position of an object in projectile motion (eg. under constant acceleration from gravity) is
\(\displaystyle x_t = x_0 + tv_0 + \frac{1}{2}t^2a\)
In this case, we have the following information:
\(\displaystyle x_0 = 3.5\)
\(\displaystyle x_4 = 0\)
\(\displaystyle a = g = -32\) (rounded off for ease of calculation)
By evaluating the position equation at t = 4, you can find \(\displaystyle v_0\). Remember that this is the vertical velocity, not the horizontal velocity.
With \(\displaystyle v_0\), you have enough information to make a parabolic equation, which you can use to get the maximum height in several ways. Please let us know if you need help from here.