Help with projectile motion question? (1 Viewer)

[sarkar_as]

1. A projectile is fired at 300 to the horizontal from the top of a cliff 200m high. Its initial velocity is 49 ms-1.

Can someone help me find its range, max height and time of flight.

Thanks.

 

[pikachu975]

I'm gonna assume you mean 30 degrees not 300.

ux = 49cos30
uy = 49sin30

Max Height:
vy = uy + ayt
0 = 49sin30 - 9.8t
t = 2.5 seconds
s = ut + 1/2 at^2
= 49sin30*2.5 - 4.9*2.5^2
= 30.625 metres is the max height

Edit: Pretty sure you have to add 200, so 230.625 metres

Time of Flight:
Let t2 be the second half of flight
s = ut + 1/2 at^2
-230.625 = 0 - 4.9t^2
t = 6.86 seconds
Therefore TOF = 6.86 + 2.5
= 9.36 seconds

Range:
Range = uxt
= 49cos30 * 9.36
= 397.19 metres

 

[sarkar_as]

wow thanks for the help

 

[sarkar_as]

For time of flight how did u know that the 2nd half of the flight isn't the same as the first?

Because I just found the time it took to reach it's max height and doubled it (so I ended up with 5).

 

[pikachu975]

For time of flight how did u know that the 2nd half of the flight isn't the same as the first?

Because I just found the time it took to reach it's max height and doubled it (so I ended up with 5).

Because it travels more vertical distance in the second half of the parabola since the first half started 200 m off the ground while the second half has to travel 200 + more.

 

[sarkar_as]

Because it travels more vertical distance in the second half of the parabola since the first half started 200 m off the ground while the second half has to travel 200 + more.

oh ok so it's only the same when the projectile starts of at 0m off the ground?

 

[jathu123]

oh ok so it's only the same when the projectile starts of at 0m off the ground?

Ye for that question, but doesn't have to be true for any other Qs.
even if it starts at 0m off the ground, it could end up landing on a 200m cliff.
The only time where you can double the time of max height to find the time of flight is when the change in y displacement is 0. ie, the projectile starts and ends on the same height.

 

[sarkar_as]

Ye for that question, but doesn't have to be true for any other Qs.
even if it starts at 0m off the ground, it could end up landing on a 200m cliff.
The only time where you can double the time of max height to find the time of flight is when the change in y displacement is 0. ie, the projectile starts and ends on the same height.

thanks