A projectile is fired upward with an initial speed vo on an airless world. A short time later, it comes back down and has a final speed vo (just before it hits the ground).

What was the sign of the total work done by the force of gravity during the flight?

a. Wgrav=0

b. Wgrav < 0

c. Wgrav >0

Explanation:

When a projectile is fired upwards with some initial speed then the it reaches the top of the projectile and then falls back to the ground.

According to the question we need to find the work done by the gravity which is acting downwards for the projectile when it is at a position just about to hit the ground in course of falling down.

As we know that work is given as:

here:

force of gravity on the object (which is acting downwards)

displacement of the object

the velocity you are looking for is made up of two parts: horizontally velocity, vh, and its vertical velocity, vv. the actual velocity of the ball will be these two parts added together, and because velocity is a vector then they will need to be added vectorially.  

neglecting air resistance etc etc.,  

the horizontal velocity component will be constant (because there is no accelerating force in the horizontal direction.  

vh = 8 ms-1  

the vertical velocity component has an accelerating force: 'g'.  

using v = u +at where a = g = 9.81 ms-2.  

the vertical component of velocity, vv = 0 + gt = vv =9.81 x 2  

vv = 19.62 ms-1 . the zero appears because the virtical component of velocity at time t=0 is zero. remember, the ball was thrown horizontally.  

how fast it will be moving (it actual path (being a parabola )) will be the vector sum of these two.  

adding vectorally gives: -  

v = sqr root (8^2 + 19.62^2) = sqr root(64+384.9) = 21.18 or 21.2 ms-1