Question

A football is kicked with an initial speed of 10.2 m/s at an angle of 40.00 above the horizontal. It lands on the ground 2.12 s later.

Determine the velocity of the football at the pinnacle of its trajectory.

Answer 1

7.66m/s

Step-by-step explanation:

Horizontal component of initial speed is

`U_x = Ucos40^0`

= 10.2 × 0.7660

= 7.66 m/s

horizontal component of the final speed at pinnacle of its trajectory is

`V_x = U_x`

= 7.66 m/s

the vertical component of the speed at pinnacle of its trajectory is

`V_y = 0`

the speed of the ball at pinnacle of the trajectory

`V = √(V_x^2 + V_y^2)`

=
`√(7.66^2 + 0^2)`

= 7.66m/s