A cannonball shot with an initial velocity of 141 m/s at an angle of 45 degrees follows a parabolic path and ?

hits a balloon at the top of its trajectory. Neglecting air resistance, how fast is the cannonball going when it hits the balloon?

• Neglecting air resistance (frictionless flight), the 141 m/s at 45 deg has a vertical component of 100 m/s and a horizontal component of 100 m/s. The vertical has gravity working against it (-9.81 m/s**2), so it will lessen to 0 as it reaches its apex. The horizontal, though has no force acting on it, so it is a constant 100 m/s.

Your answer is 100 m/s because at the top, there is no vertical velocity.

• The angle is 45 degrees. so, the x velocity and y velocity is same. at the top of the trajectory, the y velocity become 0 m/s^2. So, y velocity is the only velocity that the cannonball has when it hits the balloon. y velocity is 141/(root square of 2). Its about 100 m/s. Sorry if I make some mistake(s).

• Approximately 100 m/s in a horizontal direction. The intitial velocity in the x direction was 100 m/s as it was in the y direction. The y velocity reduces to 0 at the top of parabola, leaving only the x component.

• its velocity will b 141 * Cos35 i.e. d horizontal component of its vel. i'm assuming dat d attitude is made wid d horizontal its acc will b d acc with the aid of gravity in basic terms, as no different tension acts on it

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