CBSE Class 11-science Answered
(1) Derive equation of trajectory for horizontal projectile. 1
(2) derive the expression for time of flight & horizontal range for given parabola. 2
(3) Derive the equation for regultant velocity at any instant & also give its direction.
Asked by | 22 Oct, 2013, 01:42: PM
Expert Answer
1) The path of the projectile is shown below
The horizontal component of velocity always remain same as there is no acceleration on the horizontal axis.
Thus, the horizontal distance travelled by the projectile is
x = u cos? × t
t = x/(u cos?) ...... (1)
The vertical distance travelled by projectile is given from kinematical equation s = ut + ½at2.
Here, s = y, a = ay = -g, u = uy = u sin?
Thus, we have
y = u sin? - ½gt2 ...... (2)
Substituting equation (1) in (2), we get
This is the equation of trajectory of the projectile.
Kindly ask separate questions as separate queries.
Answered by Romal Bhansali | 24 Oct, 2013, 09:24: AM
Concept Videos
CBSE 11-science - Physics
Asked by roshnibudhrani88 | 23 Mar, 2024, 05:52: PM
ANSWERED BY EXPERT
CBSE 11-science - Physics
Asked by vinitdubey7735 | 14 Mar, 2024, 11:21: AM
ANSWERED BY EXPERT
CBSE 11-science - Physics
Asked by shailajakandikatla19 | 18 Jan, 2024, 06:40: PM
ANSWERED BY EXPERT
CBSE 11-science - Physics
Asked by dhanshreekansyakar | 09 Jan, 2024, 11:57: AM
ANSWERED BY EXPERT
CBSE 11-science - Physics
Asked by banhisikhapanda49 | 07 Nov, 2023, 10:42: PM
ANSWERED BY EXPERT
CBSE 11-science - Physics
in a vertical circle of radius r at what point in its path a particle may have tension equal to zero
Asked by momintaufik26 | 13 Oct, 2023, 08:26: PM
ANSWERED BY EXPERT
CBSE 11-science - Physics
Asked by anshujaiswalname | 10 Sep, 2023, 01:29: PM
ANSWERED BY EXPERT
CBSE 11-science - Physics
Asked by subhashreeojha235 | 20 Jul, 2023, 11:22: AM
ANSWERED BY EXPERT
CBSE 11-science - Physics
Asked by preethiprithivi05 | 21 Feb, 2023, 09:28: PM
ANSWERED BY EXPERT
CBSE 11-science - Physics
Asked by hy9022075 | 11 Jan, 2023, 05:06: PM
ANSWERED BY EXPERT