last years proj motion q (1 Viewer)

[Swalbs]

Question 16 in last years paper:
A projectile is fired at a velocity of 50 m/s at an angle of 30° to the horizontal.
Determine the range of the projectile. 4 Marks.

This is how i answered it:
v = u + at
0 = 50 -9.8t
t = 5.1 seconds
Range = horizontal velocity x time of flight
= 50cos30 x 5.1
= 220.9 metres

Is this correct? If not, how do i go about it?
Thanks in advance

 

[香港!]

Question 16 in last years paper:
A projectile is fired at a velocity of 50 m/s at an angle of 30° to the horizontal.
Determine the range of the projectile. 4 Marks.

This is how i answered it:
v = u + at
0 = 50 -9.8t
t = 5.1 seconds
Range = horizontal velocity x time of flight
= 50cos30 x 5.1
= 220.9 metres

Is this correct? If not, how do i go about it?
Thanks in advance

ur "t=5.1" is the time taken to reach max height , u need to times it by 2 to get the time of flight (ignoring air resistance)
so the 5.1 u subbed in at the end is wrong.
and also in ur "v=u+at", when u put v=0, shouldn't ur "u" be vertical component? i.e 50sin30 instead of just 50?

 

[Jago]

U_x = 50cos30U_y = 50sin30

V_y = 0 = 50sin30 - 9.8t
t_1/2 = 2.55
t_total = 5.1

therefore range = 5.1 x 50cos30 = 221

 

[word.]

you got the right answer because you coincidently did not multiply by sin30 = 1/2 for the vertical velocity or 2 to get the complete time of flight
how convenient.

 

[Swalbs]

So i made 2 mistakes, which cancelled each other out to leave me with the right answer?

 

[*yooneek*]

lol how sweet!

 

[Riviet]

Haha nice