Question

A basketball is shot with an initial velocity of 16 m/s at an angle of 55° to the
horizontal. What is the approximate horizontal distance that the ball travels in
1.5 s?

Answer 1

Answer:

approximately 13.77 m

Explanation:

We use the kinematic equations for the horizontal motion which is a uniform non-accelerated motion, and with initial velocity equal to 16 * cos(55). This is described by the equation:

$x=16 * cos(55^o) * t = 9.177 * t$

then at t = 1.5 seconds, the covered distance becomes:

x = 9.177 * (1.5) = 13.7655  m

Round the answer to the number of decimals that the problem asks.