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AS Physics HELP URGENT

I have come across a projectile motion question which I am stuck on.

A shell if fired from a gun with a velocity of 600 ms-1 at an angle of 40 degrees to the ground which is horizontal.
Calculate:
a) the time of flight
b)the range
c) the maximum height reached

ASSUME G = 10 MS-2

Can you please show answer and working.

Thank you :smile:
So the horizontal speed is 600cos40, and the vertical speed is 600sin40

a) Time Of Flight:
The time of flight will be the total time before it hits the floor again, so the final speed will equal the initial speed.

Using v = u + at (and taking UPWARDS as the positive direction)
-600sin40 = 600sin40 + (-10t)

t = 78.71 seconds

b) Distance Covered
We have the time it travels now, which is 78.71 seconds

Using S = ut + 1/2at^2
S = (600cos40 x 78.71) + 0 (Because the acceleration horizontal is 0, therefore 1/2at^2 is 0)

S = 36177 m

c) Max Height
This is when the final VERTICAL speed reaches 0.

Using v^2 = u^2 + 2as
0 = (600sin40)^2 + (2 x -10s)

S = 7437 m
Reply 2
a: t=77.1s
b: s=35453m
c: 7437m

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