help on physics?

Question

A cannonball launched with an initial velocity of 141m/s at an angle of 45degrees follows a parabolic path and hits a balloon at the top of its trajectory.Neglectin air resistance, how fast is it goin wen it hits the balloon?Wat is the acceleration of the cannonball just b4 it hits the balloon?

wat does cos mean

Answer 1

1.) Use a triangle to define the initial velocity as a horizontal velocity and a vertical velocity. The vertical velocity will be affected by gravity over time - the horizontal velocity will stay the same.

The horizontal velocity (vh) can be defined as adjacent / hypotenuse, or: initial velocity (vi) * cosine of the angle. In this case, vh = vi * cosine 45 degrees = vi * √2 / 2.

The vertical velocity (vv) can be defined as opposite / hypotenuse, or: initial velocity * sine of the angle. In this case, vv = vi * sine 45 degrees = vi * √2 / 2.

At the top of it's trajectory, the vertical velocity is 0 (because it's about to start going down instead of up. Therefore, it's total velocity = horizontal velocity = vi * √2 / 2.

The acceleration is always due to gravity, and therefore it's constantly -9.8 m/s^2. We use negative in this case as a formal way to note that acceleration goes down.

Answer 2

Simple-

Neglecting air resistance, the cannon ball will continue moving at a horizontal velocity of (141 m/s )(cos 45 degrees) = ? (I don't have a calculator)

The vertical velocity will be 0 m/s at the highest point of the trajectory.

The acceleration of the cannon ball will be the acceleration of gravity (which is negative in the upward direction) = -9.8 m/s/s.