Two dimensional motion- Projectiles

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

Time of flight, T = (2usinθ)/g

Horizontal range, R = (u²sin2θ)/g

Maximum height, H = (u²sin² θ)/2g

The range is maximum (equal to u²/g) when the angle of projection is 45˚ and the maximum height is the greatest (equal to u²/2g) 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, u²/2g = 40 and hence u²/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, u²/g = 6.The height of the wall is equal to the maximum height of the projectile so that we have (u² sin² 45˚)/2g = H. This gives H = u²/4g = 6/4 =1.5m.