Velocity and height of a ball thrown at an angle

[enantiomer1]

Homework Statement

A batted baseball leaves the bat at an angle of 35.0 degrees celcius above the horizontal and is caught by an outfielder 390 ft from home plate at the same height from which it left the bat. what was the initial speed of the ball and How high does the ball rise above the point where it struck the bat?

Homework Equations

D= v₀t
D= v₀sin 35 degrees

The Attempt at a Solution

I tried to figure it out using both the equations, but I guess I must be doing something wrong, does anyone know what equation I'm not using?

 

[buffordboy23]

First, angles do not have units in degrees celsius; angles can be in degrees or radians. You probably had a mind slip, right? =)

Second, the equations that you posted aren't useful right now, but they could be later depending on your problem-solving approach.

Let's work on the first listed unknown, the initial speed of the ball. What is the given data? We know the given angle (35 degrees), and we know that total horizontal distance the ball travels (390 feet--don't forget to convert to meters, if the answer requires it). The ball initially leaves the bat at some height that is unknown, but we do know that the ball is caught at exactly this same height. There is a specialized term to describe such a horizontal distance. What is this term? Do you have any equations for such motion that use this term?

 

[enantiomer1]

The equation I was looking for is v₀= sqrt(R[which in this case is just x]*g/sin(2 theta).
However everytime I plug it in I get sqrt(390*9.81/sin(70))= 63.81 ft/s, converted to m/s this is 63.81 ft/s * .3048 = 19.45 and this is STILL wrong... what am I not adding?

 

[buffordboy23]

Yes, as you have demonstrated, you want to use the range equation here.

Your problem is that you did not covert 390 ft into meters first. Notice that you have 390 * 9.81; 390 is in units of feet and 9.81 is in units of "meters" per second-squared. Your mixing the length units.