Finding time and range for a projectile launched from a cliff

[exi]

Traced this down to an error in my formula as I'd copied it down; it's (sqrt(Voy^2 + 2gh)+Voy) / g, not (sqrt(Voy^2 + 2gh)-Voy) / g. Thanks though!

Homework Statement

A 0.57 kg projectile is fired into the air from a cliff that's 13.9 m above a valley.
Initial velocity = 7.97 m/s
Angle: 51° above horizontal.
Acceleration = g

1: How long is the projectile in the air?
2: How far from the bottom of the cliff does the projectile land?

Homework Equations

I tried to use (sqrt(Voy^2 + 2gh)-Voy) / g to find time after breaking down the initial velocity into axial components, but this yields 1.1669 s - obviously incorrect (and verified as such).

The Attempt at a Solution

My shot at the first portion of this is above, since the equation was used in lecture as a way to find the time (in order to find the range) of an object fired at an angle from an elevated position. Just over one second doesn't check out logically, but I'm not sure why a given formula isn't yielding proper numbers.

 

[Astronuc]

This is a useful site for such problems!

http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html

 

[m2003]

I would first like to commend you for recognizing and acknowledging the error in your formula. This shows critical thinking and a willingness to troubleshoot and correct your mistakes.

To answer the first question, we can use the correct formula (sqrt(Voy^2 + 2gh)+Voy) / g to find the time of flight for the projectile. Plugging in the given values, we get (sqrt(7.97^2 + 2(9.8)(13.9))+7.97) / 9.8 = 2.39 seconds. This makes more sense intuitively, as the object should spend more than one second in the air before landing.

For the second question, we can use the equation x = Vx * t, where Vx is the horizontal component of the initial velocity and t is the time of flight. The horizontal component can be found using the formula Vx = V * cos(theta), where V is the initial velocity and theta is the angle of launch. Plugging in the given values, we get Vx = 7.97 * cos(51) = 5.18 m/s. Therefore, the distance from the bottom of the cliff where the projectile lands is x = 5.18 * 2.39 = 12.38 meters.

In conclusion, it is important to double check our formulas and calculations to ensure accuracy in our results. Keep up the good work of being a thorough and diligent scientist.