[Kinematics] Calculating the maximum height reached by the ball

In summary, the conversation discusses the use of equations to solve a problem and the importance of including a minus sign in front of the acceleration. An additional tip is provided to double check the solution by calculating the position at twice the time to maximum height.
  • #1
Slimy0233
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48
Homework Statement
A ball is thrown up at a speed of 4.0 m/s. Find the
maximum height reached by the ball. Take ##g = 10 m/s^2##
Relevant Equations
##v = u +at##
##S = ut +0.5(at^2)##
I realize I can solve the other way too. But I want to solve using the equations
##v = u +at##
##S = ut +0.5(at^2)##

and I don't know why I didn't get the right answer. Thank you for your help!
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  • #2
You forgot the minus sign in front of the acceleration.
 
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  • #3
kuruman said:
You forgot the minus sign in front of the acceleration.
ahh.... my good old archnemesis: the minus sign.

Thank you for pointing that out!
 
  • #4
An additional check to this kind of problem is to calculate the position at twice the time to maximum height. It should be zero because what goes up must come down, as they say.
 
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  • #5
kuruman said:
An additional check to this kind of problem is to calculate the position at twice the time to maximum height. It should be zero because what goes up must come down, as they say.
Thank you :smile: That's helpful. I will do that from now on.
 
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FAQ: [Kinematics] Calculating the maximum height reached by the ball

What is the formula to calculate the maximum height reached by a ball in projectile motion?

The formula to calculate the maximum height (H) reached by a ball in projectile motion is given by: \( H = \frac{v_0^2 \sin^2(\theta)}{2g} \), where \( v_0 \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity (approximately 9.81 m/s²).

How do you derive the maximum height formula?

The maximum height formula is derived from the kinematic equations. At maximum height, the vertical component of the velocity becomes zero. Using the equation \( v_y^2 = v_{0y}^2 - 2gH \) and setting \( v_y = 0 \), we get \( 0 = v_0^2 \sin^2(\theta) - 2gH \). Solving for H yields \( H = \frac{v_0^2 \sin^2(\theta)}{2g} \).

What initial conditions are needed to calculate the maximum height?

To calculate the maximum height, you need to know the initial velocity (\( v_0 \)), the angle of projection (\( \theta \)), and the acceleration due to gravity (\( g \)). These parameters allow you to use the formula \( H = \frac{v_0^2 \sin^2(\theta)}{2g} \).

How does the angle of projection affect the maximum height?

The angle of projection (\( \theta \)) significantly affects the maximum height. The height is maximized when the angle is 90 degrees (i.e., the ball is thrown straight up). For other angles, the maximum height is reduced according to the \( \sin^2(\theta) \) term in the formula \( H = \frac{v_0^2 \sin^2(\theta)}{2g} \).

Can air resistance be neglected when calculating the maximum height?

In basic kinematic calculations, air resistance is often neglected to simplify the problem. This assumption is reasonable for small, dense objects moving at moderate speeds. However, for more accurate results, especially for objects with significant air drag, air resistance should be considered, which complicates the calculations.

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