Physics question

[visionboy]

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A projectile is launched at an angle of 30° to the
horizontal and with a velocity of 100 mls. How high will
the projectile be at its maximum height?
A. 100m
B. 125 m
C. 250 m
D. 500m

Please provide explanation, thanks!

 

[SirZaitsev]

V^2 = Vi^2 + 2ad

V final will be zero since you are reaching an apex (maximal height), so you can call that 0 and solve for d (xf - xi)

Find initial velocity using trig, Vi = Vsin(30) We are concerning with the y-component of projectile since we are talking about height. We use sin for y components and cos for x. Think about the unit circle
Vi = 100 m/s * 0.5
Vi = 50 m/s
50 m/s ^2 = 2500 m/s

2500 m/s = -2(10 m/s^2)d (im rounding 9.8 m/s^2 to 10 for the gravitational acceleration since gravity is only acting on it, and we are neglecting air resistance)
Algebra at this point
-1250 m^2/s^2 = 10m/s^2 (xf - xi) (watch units since I squared the initial velocity)
-125 m = 0 - xi
We know that final position will be 0
-125 m = -xi
Cancel out the negative on both sides to get Xi = 125 m

 

[visionboy]

thanks