Question

A projectile moves from the ground such that its horizontal
displacement is x = Kt and vertical displacement is
y = Kt(1-at), where K and a are constants and t is time.
Find out total time of flight (T) and maximum height
attained (max) its
[MP PET 2008)
2K
(a) T=a, Y.
(b) T
PIMAX
(c) T=– ,
.
(d)
T=-​

Answer 1

Explanation:

y=kt-kat^2

y=ax-bx^2 form

T=rootof 2a^2/bg

$\sqrt{2a {}^{2} } \div \sqrt{b \times 10}$

=

$\sqrt{2k {}^{2} } \div \sqrt{ka10}$

$\sqrt{k \div {5a} }$

similarly

H=a^2/4b