To find the speed of the pebble when it hits the ground, we need to calculate the horizontal and vertical components of the velocity when it reaches ground level.

Step 1: Determine the Time it Takes to Hit the Ground

The height of the pebble is given by the function: $h(t) = 35t - 4t^2$ The pebble hits the ground when $h(t) = 0$: $35t - 4t^2 = 0$ This can be factored as: $t(35 - 4t) = 0$ So, $t = 0$ (when it was launched) or $35 - 4t = 0$: $t = \frac{35}{4} = 8.75 \text{ seconds}$ Thus, the pebble hits the ground at $t = 8.75$ seconds.

Step 2: Calculate the Vertical Velocity Component

The vertical component of velocity $v_y(t)$ is the derivative of $h(t)$: $v_y(t) = \frac{d}{dt}(35t - 4t^2) = 35 - 8t$ At $t = 8.75$ seconds: $v_y(8.75) = 35 - 8 \times 8.75 = 35 - 70 = -35 \text{ m/s}$ (The negative sign indicates that the pebble is moving downward.)

Step 3: Calculate the Horizontal Velocity Component

The horizontal component of velocity $v_x$ remains constant throughout the flight because there are no horizontal forces acting on the pebble (assuming we neglect air resistance). Given the initial speed of 30 m/s, the horizontal component is: $v_x = 30 \text{ m/s}$

Step 4: Calculate the Resultant Speed when the Pebble Hits the Ground

Now, the speed of the pebble when it hits the ground is the magnitude of the resultant velocity, given by: $v = \sqrt{v_x^2 + v_y(8.75)^2}$ Substituting the values: $v = \sqrt{(30)^2 + (-35)^2} = \sqrt{900 + 1225} = \sqrt{2125} \approx 46.1 \text{ m/s}$

Answer

The speed of the pebble when it hits the ground is approximately