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**Projectile**: a body that is projected by external force and continues in motion through its own inertia.

The concepts of **Newton's First Law**, and **acceleration due to gravity** both apply to **projectile motion**, and must be understood and used to complete these types of questions. If an unbalanced force acts on an object, it will accelerate in the direction of the unbalanced force. If an object is given a certain initial velocity, and there are no unbalanced forces acting on it, the object will continue along its original path in a straight line. With a projectile there is the unbalanced force of gravity which acts only downward on the object. If air resistance is negligible, there is no force acting horizontally. So, the projectile will travel horizontally with constant velocity and at the same time accelerate downward.

A shell being fired into enemy territory is an excellent example of **projectile motion** in action. The controller in charge of targeting the enemy must use radar technology to take an electronic reading of the distance. He must then use the velocity of the shell (which is a constant), and set the angle of the barrel using the distance desired. He must take into account the force of gravity as a negative acceleration and apply his knowledge of downwards acceleration due to gravity in order to succeed in hitting his target. This is an example of projectile motion, which applies the concepts of Newton's First Law of motion and the Acceleration due to Gravity concepts.

George hits a golf ball horizontally off a 25.0 m high cliff with a speed of 45 m/s. How far from the base of the cliff will the golf ball strike the ground?

*Answer* First, find time:

Vertical: d =vi (t) + (1/2)(a)(t^{2})

25 = 0 (t) + (1/2)(9.8)(t^{2})

25 = 4.9 (t^{2})

t^{2} = 25/4.9

t^{2} = 5.10

t = 2.26 sec

Find horizontal distance:

Horizontal: d =vi (t) + (1/2)(a)(t^{2})

d = (45) (2.26) + (1/2)(0)(t^{2})

d = 45 x 2.26

d = 102 m

The ball strikes the ground 102 m from the base of the cliff.

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