Question

A cannonball is fired at a 45.0° angle and an initial velocity of 670 m/s. Assume no air resistance. What is the vertical component of the cannonball’s velocity? What is the horizontal component of the cannonball’s velocity?
473.8 m/s; 473.8 m/s
-525.2 m/s; 435.5 m/s
0 m/s; 670 m/s
-378 m/s; 378 m/s

Answer 1

Answer:

473.8 m/s; 473.8 m/s

Explanation:

Given the initial velocity U = 670m/s

Horizontal velocity Ux = Ucos theta

Vertical component of the cannon velocity Uy = Usin theta

Given

U = 670m/s

theta = 45°

horizontal component of the cannonball’s velocity = 670 cos 45

horizontal component of the cannonball’s velocity = 670(0.7071)

horizontal component of the cannonball’s velocity = 473.757m/s

Vertical component of the cannonball’s velocity = 670 sin 45

Vertical component of the cannonball’s velocity  = 670 (0.7071)

Vertical component of the cannonball’s velocity  = 473.757m/s

Hence pair of answer is 473.8 m/s; 473.8 m/s