Calculating the Velocity of a Baseball Hit at 300° Angle

[Karol]

Homework Statement

First i apologize for my english, i don't know the customary names.
A baseball player hits the ball at an angle of 30⁰ above the horizon. it lands 100 meters away at the same height (another player catches it).
What is the velocity at the beginning?

Homework Equations

Constant gravitational acceleration: V=V₀-gt

The Attempt at a Solution

Total time: the time needed for going up and down (i separate the movements to horizontal and vertical):
$$0=V_0\sin30^0-gt \rightarrow t_{tot}=gV_0$$
Horizontal movement:
$$100=V_0\cos30^0\cdot t=0.87V_0\cdot 10V_0$$
$$\rightarrow 100=8.7V_0 \rightarrow V_0=3.4$$
The result should be 10 times faster: 34 m/sec

 

[goraemon]

Homework Statement

First i apologize for my english, i don't know the customary names.
A baseball player hits the ball at an angle of 30⁰ above the horizon. it lands 100 meters away at the same height (another player catches it).
What is the velocity at the beginning?

Homework Equations

Constant gravitational acceleration: V=V₀-gt

The Attempt at a Solution

Total time: the time needed for going up and down (i separate the movements to horizontal and vertical):
$$0=V_0\sin30^0-gt \rightarrow t_{tot}=gV_0$$
Horizontal movement:
$$100=V_0\cos30^0\cdot t=0.87V_0\cdot 10V_0$$
$$\rightarrow 100=8.7V_0 \rightarrow V_0=3.4$$
The result should be 10 times faster: 34 m/sec

How did you go from:
$$0=V_0\sin30^0-gt$$

to the following:
$$\rightarrow t_{tot}=gV_0$$

for the vertical component?

 

[BvU]

$0=V_0\sin30^0-gt \rightarrow t_{tot}=gV_0$

is strange. How do you do that and what do you calculate? Your relevant formula would say: that is where V = 0 in the vertical direction.

 

[Karol]

Yes, the time till V=0, double, is the time to go up and down again

 

[BvU]

Good. It is less confusing if you actually mention that extra step.
Now if V0 - gt = 0 then V0 = gt so what is T total ?

 

[Karol]

Thanks very much, i solved it thanks to Goraemon!