Question

19. The height y and distance x along the horizontal plane of projectile on a certain
planet ( no atmosphere) are given by:
y= (8t=5
and x 60
where t is in seconds
Find (a) the velocity with which the projectile was projected ?
(b) the angle with the horizontal at which projectile was projected?
(c) Find the planet on which this projectile was launched?​

Answer 1

Answer:

x=6t,u

x

=

dt

dx

=6

y=8t−5t

2

,v

y

=

dt

dy

=8−10t

v

y(t=0)

=8ms

−1

u=

u

x

2

+u

y

2

=

6

2

+8

2

=10ms

−1