Master Projectile Motion Problems with Expert Tips and Solutions

[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.

 

[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

 

[sankalpmittal]

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.

You found out ,

T= 25/v_ox

v_ox = 14

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

For 18 :

You are to find v_oy

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

Other approach is that
v_ocosθ= 14
v_osinθ=z
On dividing ,
cotθ = 14/z
solve for z...

 

[Simon Bridge]

spot the 3-4-5 triangle ;)

given initial and final displacements:
y_i=x_i=0; y_f=12m; x_f=25m

from the 3-4-5 triangle:
3v_0y=4v_0x ...1

time of flight
T = 25/v_0x ...2

12 = v_0yT - gT²/2 ...3

three equations, three unknowns.

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