Posted by **dani** on Sunday, January 4, 2009 at 8:56pm.

Two projectiles are launched form ground elvel at the same angle above the horizontal, and both return to groudn level. Projectile A has a launch speed that is twice that of projectile B. What are the ratios of the maximum heights of the balls? What are the ratios of the ranges?

B probably goes higher and farther than A, but I don't know how to find the excact ratio.

- Physics -
**drwls**, Sunday, January 4, 2009 at 9:06pm
First of all, A goes farther and higher than B. That must be what you meant to say.

Since the launch angles are the same, projectile A has twice the vertical component of initial velocity of projectile B. The maximum height attained is proprtional to the square of this velocity component and is therefore four times as high for projectile A.

The range of a projectile launched with velocity V and angle A with respect to the horizontal is

2 (V^2/g) sin A cos A = (V^2/g) sin (2A)

(See if you can derive that)

Since A is the same, the range is also four times farther when V is doubled.

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