Kinematics Question 687
Question: The height $ y $ and the distance $ x $ along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by $ y=(8t-5t^{2}) $ meter and $ x=6t $ meter, where $ t $ it’s in second. The velocity with which the projectile is projected it’s
Options:
A) 8 m/sec
B) 6 m/sec
C) 10 m/sec
D) Not obtainable from the data
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ v _{y}=\frac{dy}{dt}=8-10t $ , $ v _{x}=\frac{dx}{dt}=6 $
at the time of projection i.e. $ v _{y}=\frac{dy}{dt}=8 $ and $ v _{x}=6 $
$ \therefore v=\sqrt{v _{x}^{2}+v _{y}^{2}}=\sqrt{6^{2}+8^{2}}=10\ m/s $
