This is simply calculated by Range = t x u_h, as displacement is merely the velocity multiplied by time.

NB. velocity = displacement / time, displacement = time x velocity.
 

Q: Utilising the Tang example again:

Tang is launched with a velocity of 50m/s at 45 degrees to the ground. Assuming gravity is 9.8 m/s², how far away from his launching point will he land?

A:

The question is of course asking for Tang's horizontal range, and so to save time we need to use the answers we discovered in the previous lessons.

Tang's time of flight = u_v / 4.9 = 50 x sin 45 / 4.9 = 7.215 seconds (3 dec pl).

u_h = 50m/s x cos 45 = 35.355 m/s (3 dec pl).

Therefore Tang's range = u_h x t = 7.215 x 35.555 = approx 255.1 metres.

Questions

a) The total displacement covered.
b) The maximum height of the projectile.
c) The total horizontal distance covered.
d) Time taken for the projectile to reach the ground.

2) Range is the product of:

a) Time of flight and height.
b) Time of flight and horizontal velocity.
c) Horizontal velocity and Vertical velocity.
d) Vertical velocity and Time of flight.

3) A mango is thrown at a velocity of 25m/s at an angle of 40 degrees to the ground. Calculate the horizontal range of the mango.

4) A brick is thrown with a velocity of 20m/s at 30 degrees above the horizontal. Find the range of the brick.

5) A shell is fired with a velocity of 350m/s at 20 degrees to the horizontal from a top of a cliff 50m high. Calculate the range.

Answers

1) C
2) B
3) 62.807
4) 45.186
5) 8222.310

Now that you've learnt about the different velocity components, how to find the total time, maximum height, and range of flight, you're ready to see several harder practical examples of how they can be used to answer projectile motions questions. These examples call on all aspects of your accumulated knowledge on projectile motion, but if you've made it here so far, you shouldn't have any trouble understanding how the examples are solved.