projectile question (1 Viewer)

[Da_Bomb]

projectile is fired at 400m/s and rises for 20.4s.

(a) the initial velocity of the projectile ( shouldn't it be zero as it is at rest first--> please explain).

(b) its initial horizontal velocity .

(c) its initial vertical velocity.

(d) its range

(e) its max. height.

(f) time taken to reach max height.

(g) time of flight.

 

[smasherX21]

The initial velocity would be 400ms^-1

Then you just need to split it into componenets of horizontal and vertical velocity

 

[shady145]

The initial velocity would be 400ms^-1

Then you just need to split it into componenets of horizontal and vertical velocity

have to give angle aswell, velocity = speed+direction

 

[youngminii]

Okay in Physics (and 3u Maths), the initial time (time = 0) actually refers to the time at which it is launched.
Now someone like jm or k03### might come in here and try to argue otherwise, but for HSC purposes that will suffice.

a) Anyway, the initial velocity is 400m/s at an angle theta

b) Initial horizontal velocity = 400cos(theta), we aren't given the angle of launch..

c) Initial vertical velocity = 400sin(theta)

d) Range = UxT (Ux is initial horizontal velocity)

Now, assuming that the projectile is launched from ground level, the time of flight will end up as 40.8s
This is because all projectiles follow a parabolic trajectory, meaning its maximum height is in the middle of its flight.
Now, the question says the projectile rises for 20.4s, meaning that it reaches the maximum height after 20.4s, so we just multiply that by 2 and we have our time of flight.

Hence, range = 400cos(theta) * 40.8
= 1920cos(theta)

e) Max height is when Vy (Vertical velocity) = 0
Now, (Vy)^2 = (Uy)^2 + (2ay * height)
Therefore, 0 = (400sin(theta))^2 - (19.6 * height), this is because ay = -9.8 (gravity acts downwards, projectile motion is almost ALWAYS like this unless the question is based on a different planet etc.)
Height = [(400sin(theta))^2]/19.6

f) Time taken to reach max height is 20.4s, given in the question

g) TOF is 40.8s, as the motion is parabolic