Maximum height of launched ball

In summary, the ball was launched at an angle of 28 degrees with an initial velocity of 53m/s. It took 1.6 seconds for the ball to hit the ground when dropped straight down. The maximum height of the ball can be found by using the equation Yf = Yi + Viy(t) + (1/2)(g)(t^2) and Viy=(Vi)(sin@) to find the time it takes for the ball to reach halfway between the two points where it is 12.5 meters off the ground. This results in a maximum height of 259.25m.
  • #1
Entropia5
2
0

Homework Statement


A ball is launched at an angle of 28 degrees off a platform. The initial velocity is 53m/s. When dropped straight down, the ball hits the ground after 1.6 seconds.
What is the maximum height of the ball?

Homework Equations


Xf = Xi + Vi(t) + (1/2)(a)(t^2)
Vix = (Vi)(cos@)

The Attempt at a Solution


I was able to figure out the height of the platform from which the ball was launched, using the the 1.6s times gravity, to get 12.5m high.

I figured out that the ball traveled 5.54 seconds by solving for t in the equation Yf = Yi + Viy(t) + (1/2)(g)(t^2) and Viy=(Vi)(sin@). Using this, I found that the ball landed 259.25m away.

I was thinking I could use the equation of the y/x-graph to find the two X-values when Y = 0, and find Y at the middle X point, to get the highest point of the ball.
Which equation exactly represents this y/x-graph line? Any other suggestions would be highly appreciated!
 
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  • #2
You've pretty much got it, though there's no need to graph.

Initial velocity [itex]v_i_y = v_i sin \theta[/itex]. Due to the ball's path's symmetry, maximum height will be reached exactly halfway, temporally and spatially, between the two points where the ball is 12.5 meters off the ground. So plug in this value for [itex]Y_f[/itex] and [itex]Y_i[/itex] and calculate how much time it takes for the ball to complete this portion of its path. Then just keep in mind that maximum height is reached in half this time.
 
  • #3
swuster said:
maximum height will be reached exactly halfway, temporally and spatially, between the two points where the ball is 12.5 meters off the ground.

Of course!
Thank you for your prompt assistance :)
 

Related to Maximum height of launched ball

What is the formula for calculating the maximum height of a launched ball?

The formula for calculating the maximum height of a launched ball is: h = (v2sin2θ) / 2g, where h is the maximum height, v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.

Does the mass of the ball affect its maximum height?

Yes, the mass of the ball does affect its maximum height. The heavier the ball, the higher the maximum height will be, assuming all other factors (initial velocity, launch angle, etc.) are constant.

How does air resistance impact the maximum height of a launched ball?

Air resistance can decrease the maximum height of a launched ball. As the ball moves through the air, it experiences a force in the opposite direction of its motion, which can slow it down and decrease its maximum height. This effect is more significant for lighter and less aerodynamic balls.

At what angle should a ball be launched to achieve the maximum height?

The angle at which a ball should be launched to achieve the maximum height depends on the initial velocity and the acceleration due to gravity. Generally, a launch angle of 45 degrees will result in the maximum height for a given initial velocity and gravity.

Can the maximum height of a launched ball be greater than the initial height?

Yes, the maximum height of a launched ball can be greater than the initial height, as long as the ball has enough initial velocity and is launched at an appropriate angle. This is because the maximum height is determined by the ball's velocity and launch angle, not its initial height.

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