An outfielder throws a baseball to home plate with a velocity of +15 m/s and an angle of 30 degrees. When will the ball reach its highest alttitude?

I calculated that the max height was 11.5m, and plugged that in for y in the equation : t = sqrt(2y/g). I got 1.5s for t, but I'm not sure it's the right answer.

Question

An outfielder throws a baseball to home plate with a velocity of +15 m/s and an angle of 30 degrees.  When will the ball reach its highest alttitude?

I calculated that the max height was 11.5m, and plugged that in for y in the equation : t = sqrt(2y/g).  I got 1.5s for t, but I'm not sure it's the right answer.

anonymous · Answer

@robtobey @asnaseer

anonymous · Answer

@some_someone @Jemurray3 @helder_edwin

anonymous · Answer

I disagree with the maximum height.

anonymous · Answer

How did you calculate it?

anonymous · Answer

The maximum height?

anonymous · Answer

That velocity is the total velocity, not the y-component.  You should use the y-component of velocity to calculate maximum height.

mos1635 · Answer

|dw:1368371084921:dw|

mos1635 · Answer

highest alttitude when Vy=0
Voy-gt=0
t = Voy / g........

anonymous · Answer

I calculated the max height to be 8.60969 .  I plugged everything in and got 1.32s for the time.  Is that correct?

ksaimouli · Answer

Vx=Vot  and Vy=Vot+5t^2

ksaimouli · Answer

Vox=15cos(angle) and for y=15sin(angle)