# Find the horizontal range of a cannon ball shot with an initial velocity of 105 m/s at an angle of 35 degrees with the horizontal. With what speed is the cannon ball moving when it is at the top of its trajectory path?

The cannon ball is shot with a velocity 105 m/s at an angle of 35 degrees with the horizontal. The horizontal component of the velocity is 105*cos 35 and the vertical component is 105*sin 35. If we ignore air resistance, the horizontal component remains constant. An acceleration of 9.8 m/s^2...

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The cannon ball is shot with a velocity 105 m/s at an angle of 35 degrees with the horizontal. The horizontal component of the velocity is 105*cos 35 and the vertical component is 105*sin 35. If we ignore air resistance, the horizontal component remains constant. An acceleration of 9.8 m/s^2 acts in the downwards direction that changes the vertical component of the velocity. When the cannon ball reaches the highest point, the vertical component of the initial velocity is 0.

At the highest point of its trajectory the cannon ball has a horizontal velocity of 105*cos 35 = 86 m/s

To determine the range of the cannon ball we have to determine the time it is in motion. The time taken by the ball to reach the highest point is half the time it is in motion. Use the relation v = u + a*t where v is the final velocity of a body that starts at an initial velocity u and accelerates at a for t seconds.

Here, 0 = 105*sin 35 - 9.8*t

=> t = 105*sin 35/9.8

=> t = 6.145

The time that the cannon ball is in motion is 12.29 s. In 12.29 s the horizontal distance it moves is 12.29*105*cos 35 = 1057.15 m

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