Question

The parametric equations of a curve traced by a projectile on a certain planet with no surrounding are given by : y= 8t+ 5t^2 metre and x = 6t meter, where t is in seconds. The velocity with which the projectile is projected, is: a)8m/s. b)6m/s. c)5m/s. d)10m/s​

Answer 1

Answer:

hope it's help to you

Explanation:

Answer 2

The velocity the projectile is $10m/s$.

Explanation:

$x=6t$

$\frac{dx}{dt} =6m/s$

Assuming the projectiles initial and final height are zero.

$y=8t-5t^{2}$,$y=0$

$0=8t-5t^{2}$

$0=t(8-5t)$

$8-5t=0$

$t=\frac{8}{5} =1.6s$

$\frac{dy}{dt} =8-10t$

$=8-10*0$

$v=8m/s$

$t=1.6s$

$v=8-10*1.6$

$v=8-16$

$t=0.8s$

$v=8-10*0.8\\v=8-8$

$v=0$

$v=sqrt(8^{2}+6^{2})$

$v=sqrt(100)\\v=10m/s$