# Homework Help: Projectile motion problems

1. Jun 24, 2012

### ricky23i

Can someone tell me how to do this problem? I know 16. is A, 17 is E? 3rd pick is of what i have done not even sure if right.

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Last edited: Jun 25, 2012
2. Jun 25, 2012

### PrakashPhy

horizontal velocity is constant = $v_{x}=v_0 cos(53°)$
horizontal distance covered at any time $t$ is $S_x = v_{x}× t = v_0 cos(53°)×t$
which gives time to cover $S_x( = 25) = \frac{25}{v_0 cos(53°)} =t$
initial vertical velocity $v_y = v_0 sin(53°)$
vertical distance traveled $12 = v_0 sin(53°) ×t - \frac{1}{2} gt^2$
substituting the value of $t$ and simplifying we get

$v_0 = \sqrt {\frac{g×25^2}{2(25 tan(53°)-12) cos^2(53°)}}≈20 m/s$ [taking g = 9.8m/s^2]

so $v_x = 20 × cos(53°) ≈ 12 m/s$
etc.....
you can do the rest with all the formulas

3. Jun 25, 2012

### sankalpmittal

You found out ,

T= 25/vox

vox = 14

Are you asking 18 , 19 , ....22.

For 18 :

You are to find voy

Putting vox = 14 in T= 25/vox , find numerical value of T. Then putting formula of time of flight in a projectile you can solve for vo. Then you can find voy.

Other approach is that
vocosθ= 14
vosinθ=z
On dividing ,
cotθ = 14/z
solve for z....

Last edited: Jun 25, 2012
4. Jun 25, 2012

### Simon Bridge

spot the 3-4-5 triangle ;)

given initial and final displacements:
yi=xi=0; yf=12m; xf=25m

from the 3-4-5 triangle:
3v0y=4v0x ...1

time of flight
T = 25/v0x ...2

12 = v0yT - gT2/2 ...3

three equations, three unknowns.

One more equation comes from the slope of the vy vs t graph.