Solving for Maximum Height of Ball Launched at 60 Degrees

[meagr]

Homework Statement

shot from 60 degrees 128 cm above ground, a metal ball lands 86.5 cm away.
what is max height the ball reaches?

Homework Equations

d=v1 t + 1/2 a tsquared

v2squared = v1squared + 2ad

d= 1/2(v1 + v2)t

a=(v2-v1) / t

The Attempt at a Solution

without having my initial velocity, time, i am having trouble solving for how the max height the ball will reach .
i am trying to solve the vertical displacement being the Vy and the horizontal displacement being the Vx to find out the hypotenuse of the triangle( with angels 60, 30, 90), and i am so lost. i know i have my acceleration due to gravity, and my V2y=0m/s and my angel of elevation is 60 degrees but i am so lost in figuring out max height the ball will reach because i have so many unknowns.

can anyone help?

 

[Delphi51]

Make two headings "horizontal" and "vertical". Write x = vt under the horizontal heading and the accelerated motion formulas under the vertical heading. Fill in all the numbers you have. Note that while you don't know the initial Vx or Vy, you do know their ratio, so enter Vx as an unknown and then Vy as Vx times a tangent. If you have only one unknown in one of the formulas, begin with that! If not, you'll have to use 2 or 3 of them and solve as a system of equations.

Show your equations here if you would like more help!

 

[nlightner]

To solve for the maximum height reached by the ball, we can use the equation d=v1t + 1/2at^2, where d is the vertical displacement, v1 is the initial velocity, t is the time, and a is the acceleration due to gravity. In this case, we know the initial vertical displacement (128 cm) and the final vertical displacement (0 cm), since the ball lands on the ground. We also know the initial horizontal displacement (0 cm) and the final horizontal displacement (86.5 cm). We can also determine the time it takes for the ball to reach the ground using the equation d=1/2(v1 + v2)t, where v2 is the final velocity, which is equal to the initial velocity in the horizontal direction. We can also use the equation a=(v2-v1)/t to solve for the initial velocity in the vertical direction.

Once we have all the necessary values, we can plug them into the equation d=v1t + 1/2at^2 and solve for t. Then, we can use this value of t to plug into the equation d=1/2(v1 + v2)t to solve for the initial velocity in the vertical direction. Finally, we can use this value of v1 to solve for the maximum height reached by the ball using the equation d=v1t + 1/2at^2.

It is important to note that this solution assumes that there is no air resistance and that the initial velocity is constant throughout the motion of the ball. If these assumptions do not hold, the actual maximum height reached may vary. Additionally, the angle of elevation (60 degrees) is not necessary in solving for the maximum height, as it only affects the initial velocity in the vertical direction. I hope this helps in solving for the maximum height reached by the ball.