An object of mass 0.61 kg is initially at rest. When a force acts on it for 2.9 ms it acquires a speed of 15.5 m/s. Find the magnitude of the average force acting on the object during the 2.9 ms time interval.

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

explanation:

to push or pull an object in a certain direction is known as force. a force in moving an object from one place to another.

force depends on the mass of an object. so, more is the mass of an object more will be the force required to move the object.

thus, we can conclude that the use of force to move an object is push or pull.