Finding velocity after displacement, and an angle

[spacethisride]

Homework Statement

A driven golf ball just clears the top of a tree that is 15m in height, and 30 m from the tee. The ball then lands on the green 180m from the tee.

(a) What was the initial speed imparted to the golf ball?
(b) At what angle did the golf ball leave the tee?

Homework Equations

(a) V²=V_i² + 2aΔx

The Attempt at a Solution

(a) V²=V_i² + 2aΔx
0=V_i² + 2(-9.8 m/s²)(180 m)
V_i²=3528
V_i=59.4 m/2

I feel like I might be missing something because I wasn't sure if the 9.8 was supposed to be negative or positive...

(b) tan X = 15m/30m
arctan(1/2) = X
X = 26.57 degrees

Can I make this assumption based on the information from the problem? The only thing I wasn't sure about was if I could make a straight hypotenuse from the starting point to the top of the 15m?

 

[LowlyPion]

You should be able to do better than a straight line approximation.

What if you plotted the curve of the ball flight?

Since you know 3 points on this curve you should be able to describe it as a parabola.

Now what if you chose a smaller displacement from the tee in the x direction than just the tree? And if you kept choosing smaller and smaller displacements of x from the tee what might you call that initial line of flight? That angle would most surely be a better answer wouldn't it?.

 

[alphysicist]

Hi spacethisride,

Homework Statement

A driven golf ball just clears the top of a tree that is 15m in height, and 30 m from the tee. The ball then lands on the green 180m from the tee.

(a) What was the initial speed imparted to the golf ball?
(b) At what angle did the golf ball leave the tee?

Homework Equations

(a) V²=V_i² + 2aΔx

The Attempt at a Solution

(a) V²=V_i² + 2aΔx
0=V_i² + 2(-9.8 m/s²)(180 m)
V_i²=3528
V_i=59.4 m/2

The way you have used this equation is not quite right. You need to treat the horizontal and vertical motions separately. So for example, if you wanted to write this equation for the vertical motion, it would be:

$$v_{y}^2 = v_{0,y}^2 + 2 a_y (\Delta y)$$
and similarly for the other kinematic equations. (Notice in the way you have written the equation, you have the vertical acceleration, and a horizontal displacement.)

I feel like I might be missing something because I wasn't sure if the 9.8 was supposed to be negative or positive...

(b) tan X = 15m/30m
arctan(1/2) = X
X = 26.57 degrees

Can I make this assumption based on the information from the problem? The only thing I wasn't sure about was if I could make a straight hypotenuse from the starting point to the top of the 15m?

No, it won't be a straight line. The particle follows a parabolic path; this means the particle must initially be aimed at a point above the tree, because it will fall away from the straight line path as it moves through the air.