wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A particle is projected from point O on the ground with velocity u=55m/s at angle a=tan1(0.5). It strikes at point C on a fixed smooth plane AB having inclination of 37 with horizontal as shown in figure. If the particle does not rebound, calculate coordinates of point C in reference to coordinate system as shown in the figure.
243126_c9badb4eef6c42e0b34ff1670f5c269f.png

A
(12.5m,5m)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(5m,1.25m)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
(5m,5m)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
(1.25m,1.25m)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B (5m,1.25m)
If u is the initial velocity, ucosα = horizontal constant velocity (unaffected with vertical acc. g)

usinα = vertical velocity
and t seconds. Let the co-ordinates of point C(a, b)

a = ucosαt ----- (i)

b = usinαt - 129.81t² ------ (ii)

from equations (i) & (ii) with a substitution for t we get,

b = atanα -
12(9.81)a² / [ucosα]² -------- (iii)

the equation of the line representing the incline which passes through (a, b) is y = xtan37° + c
when y = 0 => x = 3.333, therefore c = -3.333(0.7536) = -2.5118

the equation is y = 0.7536x - 2.5118

therefore b = 0.7536a - 2.5118 --------- (iv)

substituting in equation (iii),

0.7536a - 2.5118 = 0.5a -
12(9.81)a² / [5√5(0.8944)]²

- 2.5118 + 0.2536a = - 4.905a² / 100

251.185 = 4.905a² + 25.36a
a² + 5.17a - 51.21 = 0
on solving the above eqn we get,

solving a ~= 5.02 m, b ~= 1.27 m =
(5.02, 1.27) .
Thus the coordinates are (5m,1.25m)

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon