The equations you have to remember in projectile motion are the following (with usual notations):

Time of flight, **T = (2usinθ)/g**

Horizontal range, **R = (u**^{2}sin2θ)/g

Maximum height, **H = (u**^{2}sin^{2} θ)/2g

The range is maximum (equal to u^{2}/g) when the angle of projection is 45˚ and the maximum height is the greatest (equal to u^{2}/2g) when the angle of projection is 90˚. Now consider the following **M.C.Q.:**

**The greatest height to which Tom Sawyer could throw a stone is 40m. The greatest horizontal distance to which he could throw the stone is**

**(a) 40m (b) 60m (c) 80m (d) 100m (e) 160m**

We have, u^{2}/2g = 40 and hence u^{2}/g = 80. The correct option is (c). You should remember that the greatest height ‘h’ to which you can throw a stone is related to the greatest horizontal distance ‘R’ to which you can throw it as R = 2h.You should remember the following points also:

(1) There are two angles of projections for which the horizontal ranges are the same and these angles are equally spaced with respect to 45˚. For example, angles of projection 20˚ and 70˚ will yield the same horizontal range (since they are separated by 25˚ with respect to 45˚ line of projection).The above fact can be put in a different manner:If two particles are projected with the same velocity at angles θ and 90-θ, they have the same range.

(2) If H1 and H2 are the maximum heights reached in the above cases of the same range R, then** R = 4√(H1H2)**

(3) Maximum range of a projectile is 4 times the maximum height attained when the projectile has the maximum range: **R**_{max}= 4H

(4) Horizontal range R and maximum height H of a projectile are related as **R = 4Hcotθ**.

All the above results can be easily obtained using the standard expressions for horizontal range and maximum height.Let us now consider the following M.C.Q.:

**From a point on the ground at a distance 3m from the foot of a wall, a stone is thrown at an angle of 45˚. If the stone just clears the top of the wall and then strikes the ground at a distance of 6m from the point of projection, the height of the wall is**

**(a) 1.5m (b) 2m (c) 3m (d) 4.5m (e) 6m **

The correct option is (a) which you can select immediately if you remember statement (3) given above. If you don’t remember the statement you can solve the problem this way:

Since this is a case of maximum range, u^{2}/g = 6.The height of the wall is equal to the maximum height of the projectile so that we have (u^{2} sin^{2} 45˚)/2g = H. This gives H = u^{2}/4g = 6/4 =1.5m.