What is the maximum height to which the ball rises if a spring gun projects a golf ball at an angle of 45 above the horizontal? The horizontal range is 10m.

Question

anonymous · Answer

sir can you help me with this? given is kinda lack, i don't know what formula to use

anonymous · Answer

you should divide the velocity to 2 components, one of them is vertical the other is horizontal. For vertical component, Vx= V*Cos(45)  you can find the vertical component.

After that, you know the range, so  you should multiply Vx with time t.  then you get V*t 

you can use this for vertical calculation...

anonymous · Answer

still kinda dizzy so i get Vox=V*cos45 and Voy=V*sin45
what will i do with this?.. x.x

anonymous · Answer

x = v cos 45 . t
10 = ½√2 vt

then, you can use this formulas 
vy = voy - gt
vy² = voy² - 2gh
with assume that vy = 0 at maximum height

anonymous · Answer

wait ill try to solve.. ill just ask if something bothers me thanks for the help guys :)

anonymous · Answer

@surygita the formula $$10=\frac{ 1 }{ 2 }\sqrt{2}vt$$ is it from the formula $$h=vot \pm \frac{ 1 }{ 2 } g t ^{2}$$ ?

anonymous · Answer

woops, I made my mistake
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