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## Homework Statement

A ball is kicked with an initial speed of 20m/s at an angle of 50° above the horizontal. What maximum height does the ball reach?

## Homework Equations

- Vfx=VicosΘ

- Δx=VicosΘt

- Vfy=(VisinΘ)+ay⋅t

- (Vfy)^2=(VisinΘ)^2+2ay⋅Δy

- Δy=1/2(Vfy+(VisinΘ))⋅t

## The Attempt at a Solution

I know that the final velocity for both x and y is 20m/s if that is the case. The acceleration for the x-direction is 0m/s^2 while its -9.8m/s^2 due to earth gravitational acceleration. Vy at max height is 0.

**Known**- Vfx and Vfy= 20m/s
- Θ=50°
- Vfy=0m/s
- ax=0m/s^2
- ay=-9.8m/s^2
- T=?
- Δy=?
- Δx=?

I'm assuming that the question is asking for max height as Δy in the y-direction. I was trying to solve for time in order to use the equation to solve for Δy but I can't see how to do that. I'm not sure if there is another way that I could do this or if I just labeled things incorrectly.