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.

 

[asteriatic]

Based on the information provided, it seems like you have correctly calculated the time of flight and the horizontal distance traveled by the ball. However, there seems to be an error in your calculation for the vertical distance. The equation you used, d=v1t+1/2at^2, is the correct equation for calculating displacement (d) in projectile motion, but you have used the initial vertical velocity (v1) instead of the final vertical velocity (vf). The correct equation should be d=vf*t+1/2at^2. Also, it is important to note that in this equation, the value of a should be negative (-9.8 m/s^2) since it represents the acceleration due to gravity pulling the ball downwards.

Using the correct equation, we can calculate the final vertical velocity as vf= v1 + at. Plugging in the values, we get vf= 9.64 + (-9.8)(2.26) = -11.412. Now, we can plug this value into the displacement equation, d= vf*t+1/2at^2, to get d= (-11.412)(2.26) + 1/2*(-9.8)*(2.26)^2 = -2.5 m.

Since the ball lands 26 m away from the cliff, we can set up a right triangle with the horizontal distance (26 m) as the base and the vertical distance (-2.5 m) as the height. Using the Pythagorean theorem, we can solve for the height of the cliff as h=√(26^2 - (-2.5)^2) = 25.99 m, which can be rounded to 26 m.

Therefore, the height of the cliff is 26 meters. I hope this helps you understand the steps you were missing and how to arrive at the correct answer of 3.2 m. Keep practicing and you will become more confident in solving projectile motion problems. Good luck!