Question

A projectile is fired at an angle such that the vertical component of its velocity

and the horizontal component of its velocity are BOTH equal to 30 m/s.
a. Using the approximate value of g = +/-10 m/s2, how long in time does it take for the
ball to reach its high point?

Answer 1

It takes 3 seconds to reach its high point

Step-by-step explanation:

Projectile Motion

In a projectile motion (or 2D motion), the object is launched with an initial angle θ and an initial velocity vo.

The components of the velocity are

`v_(ox)=v_o\cos\theta`

`v_(oy)=v_o\sin\theta`

The speed in the horizontal direction at any time t is:

`v_y=v_o\sin\theta-g.t`

The time taken to reach the maximum height is when vy=0, or:

`\displaystyle t_m=(v_o\sin\theta)/(g)`

We are given the y-component of the velocity, thus:

`\displaystyle t_m=(30~m/s)/(10~m/s^2)`

`t_m=3~s`

It takes 3 seconds to reach its high point