Horizontal / Vertical Motion, Parabolic Motion

[Mello]

If this sounds ridiculously vague, I apologize - my physics teacher gave us this problem verbally and didn't want to clarify anything.

Homework Statement

A ball is thrown at a 45 degree angle to the ground and remains in the air for 2.35 seconds by an individual standing 1.81m. How far did the ball travel?

Homework Equations

I'm not really sure what equations would be suitable to use - I am unsure how I can calculate the initial horizontal speed of the ball. I know the horizontal and vertical components would be equal since it is a 45 degree angle but beyond this I am at a bit of a loss.

I used vyf = vyo + gt to calculate that vyf was 23.03, however I am unsure how to factor in the individuals height to the equation.

 

[SammyS]

If the individual throws a ball straight up (vertically) and it's in the air for 2.35 seconds, can you determine what the initial velocity is ?

 

[Mello]

Final vertical velocity = Initial vertical velocity + (force of gravity * time)
0 m/s = Vyo + (-9.81 m/s * 2.35s)
Vyo = ~23 m/s
Maybe?

 

[SammyS]

Final vertical velocity = Initial vertical velocity + (force of gravity * time)
0 m/s = Vyo + (-9.81 m/s * 2.35s)
Vyo = ~23 m/s
Maybe?

That's if it takes 2.35 seconds to reach it's highest point.

 

[Mello]

Ahh, so it would be half of that so ~11.5 m/s?

If Vyo is 11.5 in a straight up situation, then I could assume (since the angle is 45 degrees) that Vyo = Vxo and therefore each initial velocity would be ~5.75 m/s?

 

[SammyS]

Ahh, so it would be half of that so ~11.5 m/s?

If Vyo is 11.5 in a straight up situation, then I could assume (since the angle is 45 degrees) that Vyo = Vxo and therefore each initial velocity would be ~5.75 m/s?

No. You can assume that the vertical component, (v₀)_y is ~11.5 m/s .

Since the launch angle is 45°, the horizontal component of the velocity is also ~11.5 m/s .