Projectiles

A projectile is an object thrown into the air or launched into space and moves freely by itself under the influence of gravity and air resistance. Examples can be located in sports and warfare.

Terms in projectile explained

Time of flight (T) can be defined as the time required for it to return to the same level from which it was projected.

T= 2UsinO/g

The range (R) is defined as the horizontal distance from the point of projection to the point where the projectile hits the projection plane again.

R= U²sin2O/g

R_max= U²/g

Maximum height (H) is defined as the highest vertical distance reached as measured from the horizontal projection plane.

H= U²sin²O/2g

Note that it is advisable to work from first principles with equations y=utsinO-½ gt²

x=utcosO

Past questions

1.A particle is projected horizontally at 10m/s from the top of a tower 20m high. Calculate the horizontal distance travelled by the particle when it hits the level ground. (g= 10ms^-2) (Wassce 2014)

Answer: u= 0ms^-1, S= 20m,

S= ut + ½ gt²

20= 0×t+½ ×10×t²

20= 0+5t²

20= 5t²;   5t²=20; t²= 4; t= √4; t= 2sec

t= 2sec, u= 20m/s

Horizontal distance= Velocity×time

                              = 20×2 =40m

2. A projectile is released with a speed u at an angle O to the horizontal. With the aid of a diagram, show that the time of flight is equal to 2usinO/g, where g is the acceleration of free fall.( Wassce 2013)

Answer:

3. A particle is dropped from a vertical height h and falls freely for a time t. With the aid of a sketch, explain how h varies with: (a) t; (b)t².(Wassce 2009)

Answer:

From diagram, H= vertical distance

R= horizontal distance

So u= 0m/s, s= ut + ½ gt²

Since H= S

Then, H= 0(t)+½ gt²

H= ½ gt², 2H=gt², 2H/g = t², t= + √2H/g

4. A particle is projected horizontally at 10m/s from a height of 45m. Calculate the horizontal distance covered by the particle before hitting the ground. (g= 10m/s) (Wassce 2006).

Answer: y= utsinO-½ gt²;

45= 0×t×sinO-½ ×10×t²

45= 0-5t²; 5t² = 45; t²= 45/5=9;

t=√9=3s. Horizontal distance is given by X= utcosO= 10×3cosO= 30×1= 30m.

5. A stone, thrown horizontally from the top of a vertical wall with a Velocity of 15m/s, hits the horizontal ground at a point 45m from the base of the wall. Calculate the (I) time of flight of the stone; (g= 10m/s) (Wassce 2004)

Answer: R= ut+½ gt²; since g = 0; R= it

45= 15×t; T= 45 = 3seconds

                      15

6. State two objects each used in sports and warfare which may be considered as projectiles. (Wassce 2003)

Answer:  for sports: javelin and shot put

For warfare: spear , bow and arrow

7. A projectile is fired with a Velocity of 20m/s at an angle of 40° to the horizontal. Determine the components of the velocity of the projectile at its maximum height. (Wassce 2016)

Answer: U= 20m/s; q= 40°

Vertical component of the velocity at it maximum height is Uy= UsinO

Uy= 20sin40

     20×0.643

= 12.86m/s.

8.what is a projectile? (Wassce 2015)

Answer:A projectile is an object thrown into the air or launched into space and moves freely by itself under the influence of gravity and air resistance.