Question

The height y and horizontal distance x covered by a projectile in a time t seconds are given by the equations y=8t−5t^2 and x=6t where t is in seconds the angle at which the projectile was projected is​

Answer 1

Answer:

x=6t

on differentiating with respect to t

v

x

=6

y=8t−5t

2

v

y

=8−5t

at t=0 v

y

=8

v

2

=6

2

+8

2

=10

2

v=10m/s

Answer 2

ANSWER

v=10m/s

Explanation:

x=6t

on differentiating with respect to t

v

x

=6

y=8t−5t

2

v

y

=8−5t

at t=0 v

y

=8

v

2

=6

2

+8

2

=10

2

v=10m/s