Question

The height y and distance x along the horizontal for a body projected in the xy plane are given by y=8t-5t^2 and x= 6t . The initial speed of projection is ?.

Answer 1

y = 8t - 5t²
differentiate wrt time
dy/dt = 8 -10t

we know, change in position per unit time is called , velcity .
e.g dy/dt = Vy

Vy = 8 -10t
now at t = 0 ( initially )
Vy =8 -10× 0 = 8

x = 6t
differentiate wrt time
dx/dt = 6
similarly , dx/dt = Vx =6

now,
V = Vx i + Vy j
= 8 i + 6 j

magnitude of velocity =| V | =√(8² + 6²) =10 m/s

Answer 2

Answer:

10 m/s

Step-by-step explanation:

y = 8t - 5t²

differentiate wrt time

dy/dt = 8 -10t

we know, change in position per unit time is called , velcity .

e.g dy/dt = Vy

Vy = 8 -10t

now at t = 0 ( initially )

Vy =8 -10× 0 = 8

x = 6t

differentiate wrt time

dx/dt = 6

similarly , dx/dt = Vx =6

now,

V = Vx i + Vy j

= 8 i + 6 j

magnitude of velocity =| V | =√(8² + 6²) =10 m/s