The height y and the distance x along the . . .

Question : The height y and the distance x along the horizontal, for a body projected in the vertical plane are given by y = 8t – 5t² and x = 6 t. What is initial velocity of the body?

Doubt by Krish

Solution : 

During Horizontal Motion 
x=6t (Given)

We know, 
v_x=dx/dt

v_x=d(6t)/dt

v_x=6 m/s

For initial velocity, t=0 sec

v_x at t=0

v_x=6 m/s

During Vertical Motion 

y = 8t – 5t² (Given)

We know, 

v_y=dy/dt

v_y=d(8t-5t2)dt

v_y=8-10t

For initial velocity, t=0 sec

v_y at t=0

v_y=8-10(0)

v_y=8-0

v_y=8 m/s

The total velocity of the body will be given by the resultant of these two velocities. 

Since these horizontal and vertical velocities are perpendicular to each other so the angle between the two will be 90° i.e. θ=90°

v=√[v_x²+v_y²+2v_xv_ycosθ]
v=√[(6)2+(8)2+2(6)(8)cos90°]

v=√[36+64+96×0]
v=√[100+0]
v=√100
v= 10 m/s

Hence, the initial velocity of the body is 10 m/s.