projectile question (1 Viewer)

[wolf7]

hey, i got to projectile qiuestion

question 1

An inexperienced shooter aims a rifle horizontally at a target 300m way. if the bullet leaves that barrel of the gun at 2500m/s, how far below the centre of the target will the bullet strike

question 2

A bullet fired horizontally at a target 250m away strikes the target 15.0cm below the centre. with what velocity did the bullet leave the gun

 

[helper]

1. Calculate your time of flight using your initial horizontal velocity and range.
Then calculate how far it drops from this time and your vertical information.

2. Make an expression for range in terms of u (not u_x) and t. Rearrange to make t the subject of the formula.

Do a similar equation for the vertical. Substitute, your expression for t from the horizontal into the vertical.

 

[wolf7]

i got the working out bit, u reakon u cacn telll us answer becuase teh book doesnt show the answer

 

[wolf7]

also can u show the working out

 

[simbim21]

i worked it out and got:
1. Range=horizontal component x time
300=2500 x t
therefore, t=0.12sec
using s=ut+1/2at(squared) - vertical components of u=0, a=9.8
= o + 1/2(9.8)(0.12squared)
= 7.056 cm

2. s=ut+1/2at)squared)
0.015= 0 + 1/2(9.8)tsquared
therefore, t= 0.055 sec
R=ut
250=u x 0.055
therefore u= 4518.5 m/s

hope that helped! (i hope its right , i only just started yr.12)

 

[jono_z1]

2. s=ut+1/2at)squared)
0.015= 0 + 1/2(9.8)tsquared
therefore, t= 0.055 sec
R=ut
250=u x 0.055
therefore u= 4518.5 m/s

isn't distance supposed to be 0.15m (=15cm) rather than 0.015m (=1.5cm)?

r=ut+.5at^2
0.15 = 0 + .5 x -9.8 x t^2
t = 0.17s

r=ut
250=u x 0.17
u=1428.87 m/s (i only started yr 12 as well, so i could be wrong too)

 

[Riviet]

isn't distance supposed to be 0.15m (=15cm) rather than 0.015m (=1.5cm)?

r=ut+.5at^2
0.15 = 0 + .5 x -9.8 x t^2
t = 0.17s

r=ut
250=u x 0.17
u=1428.87 m/s (i only started yr 12 as well, so i could be wrong too)

That's all correct. I got the same answers as you.