Question

A cannon fires a cannonball 500.0 m downrange when set at a 45.0o angle. At what velocity does the cannonball leave the cannon?

Answer 1

Answer:

v = 70 m/s

Explanation:

Range on the cannon ball is given as

$d = 500.0 m$

here the angle of the projection of the ball is given as 45 degree

now we know that if the velocity of the ball is "v" then its two components will be given as

$v_x = vcos45$

$v_y = vsin45$

so here time of flight of the motion is given as

$T = \frac{2v_y}{g}$

$T = \frac{2vsin45}{g}$

also the range is given as

$R = v_x T$

$R = (vcos45)(\frac{2vsin45}{g})$

now plug in all data in this equation

$500.0 = \frac{v^2(2sin45cos45)}{g}$

$v = 70 m/s$