Amy's Ball: Solving a Cliff-side Problem

[scottyourban]

Homework Statement

alright. story time.
Amy throws a ball at 12 m/s at an angle of 23 degrees above the horizontal, off a cliff that is 35 m high. How far from the base of the cliff should Amy start looking for the ball?

Homework Equations

V_yi=V_i(sin θ)

t=V_yi - V_y / g
Y_max=y_i + v_yi + .5at²

The Attempt at a Solution

well, first i took 12*(sin23) which gave me 4.69 (rounded).
then, i found the time: so i took (12-0)/9.8, which gave me 1.22 seconds
then, normally, to find the distance, i would take the square root of (144-4.69²)

BUT, the problem is, what do i do to continue it? those equations would work if i was finding where it landed at a vertical distance that is = to where the ball started it, but we have to factor in that extra 35 m. so, my question is, how do i calculate all that in?

 

[IBY]

Hold on, shouldn't the formula for time be a quadratic formula? The time it takes for the ball to land on the ground depends on three terms: gravitational acceleration, height of the cliff, and the launch velocity.

Anyways, with time, use it in the kinematic equation in the x direction.

 

[PhanthomJay]

Homework Statement

alright. story time.
Amy throws a ball at 12 m/s at an angle of 23 degrees above the horizontal, off a cliff that is 35 m high. How far from the base of the cliff should Amy start looking for the ball?

Homework Equations

V_yi=V_i(sin θ)

t=V_yi - V_y / g
Y_max=y_i + v_yi + .5at²

The Attempt at a Solution

well, first i took 12*(sin23) which gave me 4.69 (rounded).

OK, that gives you the initial y component of the velocity

then, i found the time: so i took (12-0)/9.8, which gave me 1.22 seconds

by setting vy =0, you are finding the time it takes to reach the maximum height

then, normally, to find the distance, i would take the square root of (144-4.69²)

That gives you the horizonatl component of the velocity (which is more easily found from Vcostheta)...that doesn't give you the distance the ball travels before it hits the ground. Try first solving for the time it takes to hit the ground using your third equation, except instead of using ymax, what is the value of y when the ball hits the ground?

EDIT: As per IBY's suggestion...I posted a bit late!

 

[ashishsinghal]

"then, i found the time: so i took (12-0)/9.8, which gave me 1.22 seconds
then, normally, to find the distance, i would take the square root of (144-4.69²)"what have you done here? - it is not even the time taken to reach the highest point...
that would be 4.69/9.8 . the time taken to reach the ground would be found using formula
s = ut + 0.5at*t

 

[ashishsinghal]

well, first i took 12*(sin23) which gave me 4.69 (rounded).
then, i found the time: so i took (12-0)/9.8, which gave me 1.22 seconds
then, normally, to find the distance, i would take the square root of (144-4.69²)
?

when you have found out the time using the formula above then distance would be found this way,
multiply horizontal velocity to the time

 

[rubyrubyahaha]

Range=cos(angle)*V*t

Using quadratics, one can derive the formula for t that includes the original height (h):

t=[(2*h*g+(sin(angle)V)^2)^0.5)+sin(angle)V]/g

(I can demonstrate how I derived the above if wanted)

Therefore:

Range=cos(angle)*V*[(2*h*g+(sin(angle)V)^2)^0.5)+sin(angle)V]/g

Which, for your particular problem, should yield approximately 35m.