Catapulted

Calculus Level 3

On NLLM (a satellite of the planet VZIGS) the gravitational acceleration is exactly $3 \text{ m/sec}^2$ .

A catapult launches a stone from ground level at a velocity of $13 \text{ m/sec}$ . The horizontal component of the velocity is exactly $5 \text{ m/sec}$ . At the peak of the stone's trajectory, what is its displacement from its starting point (in $\text m$ )?

The answer is 31.241.

1 solution

Denton Young
May 19, 2016

The initial upward component of the velocity is $\sqrt(13^2 - 5^2)$ = 12 m/sec.

Therefore, the stone reaches its peak at 12/3 = 4 sec into the flight. At this time the horizontal displacement is $4\times 5$ = 20 m.

The vertical displacement: accel = -3 $m/sec^2$ , so velocity = 12 - 3t $m/sec$ and position =12t - (3/2)t^2 m

For t = 4, this is 24 m.

So total displacement = $\sqrt(20^2 + 24^2)$ = $\sqrt(976)$ = 31.241 m

Moderator note:

Simple standard approach.

Catapulted

The answer is 31.241.

1 solution

Moderator note:

0 pending reports