Question

A projectile is launched horizontally from a cliff top at 12 m/s. Determine the x-y positions at 1-second intervals. The launch position is (0 m, 0 m).

Answer 1

The x position will be "12 m" and the y position will be "-4.9 m".

According to the question,

Along x direction, Initial velocity:

• 12 m/s

Along x direction, Acceleration is:

• 0

Along y direction, Acceleration is:

• -10 m/s²

Time,

• t = 1 second

→ The X coordinate will be:

= $12\times t$

= $12\times 1$

= $12 \ m$

→ The Y coordinate will be:

= $-(u_y t+\frac{1}{2}a_y t^2 )$

= $0+4.9 t^2$

= $4.9 \ t^2$

= $-4.9\times 1^2$

= $-4.9 \ m$

Thus the above answer is correct.

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