Assistance with projectile motion question

[gbaby370]

I had recently received an exam back rom my kinematics unit in my physics course, this is the only question that I answered and was incorrect. I was wondering if someone could assist me on what steps I am missing. The correct answer is 3.2m, but I can't figure out how to get to it.


A ball is launched from a cliff with and initial velocity of 15 m/s at an angle of 40o above horizontal. If the ball lands 26 m away, determine the height of the cliff.

Horizontal

V=cos40(15)
v=11.5

T=d/v
T=26/11.5
T=2.26s

Vertical

V1=sin40(15)
v1=9.64

d=v1t+1/2at^2
d=9.64(2.26)+1/2(-9.8)t^2
d=-2.5m

Any feedback would be greatly appreciated.

 

[e^(i Pi)+1=0]

You made some kind of simple arithmetic mistake in the final equation.

9.64(2.26)+1/2(-9.8)(2.26)²≠-2.5

 

[jing2178]

d=9.64(2.26)+1/2(-9.8)t^2

Any feedback would be greatly appreciated.

I get -3.24 for this calculation using t=2.26

 

[gbaby370]

I figured out that it was a matter of simply not taking my time. Thanks for the help, greatly appreciated.