An object is launched into the air a with an initial velocity of 50ms-1 at and angle of 30o.
i) ux = uCos(Θ) = 50Cos(30) = 43.30 ms-1
x = uxt = 43.30 x 3 = 129.90 m
ii) uy = uSin(Θ) = 50Sin(30) = 25 ms-1
y = uyt - 1/2gt2
a) y = 25 x 2 - 1/2 x 9.81 x 22 = 30.38m
b) y = 25 x 3 - 1/2 x 9.81 x 32 = 30.86m
c) y = 25 x 4 - 1/2 x 9.81 x 42 = 21.52m
ii) vy = uy - gt
a) vy = 25 - 9.81 x 2 = 5.38 ms-1
b) vy = 25 - 9.81 x 3 = - 4.43 ms-1
c) vy = 25 - 9.81 x 4 = - 14.24 ms-1
The diagram above summarises the results.
The green arrow represents the horizontal distance travelled at 3 seconds
The blue arrows represent the heights at 2,3, & 4 seconds
The red horizontal arrows represent the constant horizontal velocity of the projectile
The vertical red arrows show the changing magnitude and direction of the projectile's vertical velocity at 2,3 & 4 seconds.
Assuming that the projectile travels above level ground, calculate;
a) the maximum height the projectile reaches
b) the horizontal distance it travels before it hits the ground
a) vy = uy - gt
At maximum height vy = 0
Therefore t = uy/g
t = 25/9.81 = 2.55 s
y = uyt - 1/2gt2
At 2.55 seconds
y = 25 x 2.55 - 1/29.81 x 2.552 = 31.86m
b) It takes 2.55s to reach maximum height so it will take another 2.55 s to hit the ground
Therefore the projectile is in flight for 5.1 seconds
x = uxt
x = 43.30 x 5.1 = 220.83m
In the this question we made use of the fact that the time it would take for the object to drop back to the ground from its max height would be the same as the time taken for it to initially reach this height after it left the ground. This is always true when an object is fired vertically upwards (ignoring air resistance)as it lands at exactly the same point it was launched from. It is also true for a projectile if the ground it travels over is level. However this is not always be the case, the ground where the projectile lands could be higher or lower than where it was launched!