Example. John throws a frisby off the top of a hill 20m high, he throws it with a speed of 8.6m/s, at what speed will the frisby hit the ground below the hill?
John throws a frisby off the top of a hill 20m high, he throws it with a speed of 8.6m/s, at what speed will the frisby hit the ground below the hill? John’s hand is 1.5m above the surface of the hill.
At first glance it seems that we are given an insufficient amount of data but by combining our knowledge of uniform acceleration and projectile motion we will be able to do solve this question.In the diagram above you can see John performing the aforementioned actions. The frisby starts off 20 + 1.5 metres above the ground (hill + john’s hand’s height), it has a velocity to the right and it has a velocity down. The horizontal velocity is given to us, and the frisby will maintain it for the whole duration of the motion. The downward velocity is not constant, it accelerates constantly with a magnitude of 9.8m/sІ [down] and we will have to use the uniform acceleration formulae to find out the downward speed of the frisby upon impact. The blue vector is the resultant velocity, the magnitude of which we need, and is found by adding the two black vectors together. The green line is the frisby’s physical path.
Lets list the data that we know, along with the direction ( horizontal/vertical ) it is associated with.
Lets consider the right direction and the downward direction to be positive. The vertical and horizontal components of motion can be related to each other by the time. Since the time of the motion is the same in both directions we can find it using the horizontal data we know.d = 1/2 (Vfx + Vix) Ч t
t = 2d / ( Vx + Vx )
t = 2( 21.5 ) / 2( 8.6 )
t = 21.5 / 8.6
t = 2.5s
Now that we know that the whole motion took 2.5 seconds we can use uniform acceleration formulae to find out Vfy.
a = (Vfy – Viy) / t
Vfy = ta + Viy
Vfy = (2.5)(9.8) + 0
Vfy = 24.5m/s
Now that we know that the components of the final velocity are 8.6m/s [right] and 24.5m/s [down] we can use the pythagorean theorem to find the impact speed of the frisby saucer:
VimpactІ = 8.6І + 24.5І
VimpactІ = 73.96 + 600.25
VimpactІ = 674.21
Vimpact = 25.96 m/s
We conclude that when the frisby lands it has a speed of 26m/s . That’s quite a gain from the starting speed of 8.6m/s, all of that extra speed is due to gravity.
Lets convert the units of the impact speed into kilometres per hour to give us an idea of the frisby’s fun level.
(26m/s Ч 3600) / 1000 = 93.6 km/h This frisby is as fast as a car, and will probably create a crater in the ground.