Question

If the instantaneous velocity of a particle projected
as shown in figure is given by y = a+(b-ct)ſ
where a, b and care positive constants, the range
on the horizontal plane will be :
y A
V
X
(1) 2ab/c
(3) ac/
(2) ab/c
(4) a/2bc​

Answer 1

Given : If the instantaneous velocity of a particle projected as shown in figure is given by $\vec v=a\hat i + (b - ct)\hat j$

To find : The range on the horizontal plane will be..

solution : instantaneous velocity of a particle is given by,$\vec v=a\hat i + (b - ct)\hat j$

at t = 0, initial velocity, $u=a\hat i + b\hat j$

differentiating v with respect to time we get, acceleration of particle, $a=0\hat i - c\hat j$

so vertical component of acceleration, $a_y=-c$

vertical component of velocity, $u_y=b$

To find horizontal range,

vertical component of Particle = 0

⇒$u_yt+\frac{1}{2}a_yt^2=0$

⇒bt + 1/2 × (-c)t² = 0

⇒t = 2b/c .....(1)

now horizontal component of displacement of particle at time t = 2b/c = horizontal range

= $u_xt+\frac{1}{2}a_xt^2$

= a × 2b/c + 0

= 2ab/c

Therefore the horizontal range of particle is 2ab/c

Answer 2

Answer:

check the answer

thanks