15 Sy A dynamite blast at a quarry launches a chunk of rock straight upward, and 2.0 s later it is rising at a speed of \( 15 \mathrm{~m} / \mathrm{s} \). Assuming air resistance has no effect on the rock, calculate its speed: at launch; 5,0 s after launch.

To solve this problem, we can use the kinematic equations for uniformly accelerated motion. We'll assume upward is the positive direction and take the acceleration due to gravity \( g = 9.8 \, \text{m/s}^2 \) (acting downward). ### 1. **Speed at Launch** Let \( v_0 \) be the initial velocity (speed at launch). The velocity of the rock after time \( t \) is given by: \[ v = v_0 - g t \] We know that \( v = 15 \, \text{m/s} \) at \( t = 2.0 \, \text{s} \). Plugging in the values: \[ 15 \, \text{m/s} = v_0 - (9.8 \, \text{m/s}^2)(2.0 \, \text{s}) \] \[ 15 \, \text{m/s} = v_0 - 19.6 \, \text{m/s} \] \[ v_0 = 15 \, \text{m/s} + 19.6 \, \text{m/s} = 34.6 \, \text{m/s} \] **Initial Speed at Launch:** \( \boxed{34.6 \, \text{m/s} \text{ upward}} \) ### 2. **Speed 5.0 Seconds After Launch** Using the same equation for \( v \) at \( t = 5.0 \, \text{s} \): \[ v = 34.6 \, \text{m/s} - (9.8 \, \text{m/s}^2)(5.0 \, \text{s}) \] \[ v = 34.6 \, \text{m/s} - 49 \, \text{m/s} = -14.4 \, \text{m/s} \] The negative sign indicates the rock is moving **downward** at this time. **Speed at 5.0 Seconds After Launch:** \( \boxed{14.4 \, \text{m/s} \text{ downward}} \) ### **Summary** - **At Launch:** \( 34.6 \, \text{m/s} \) upward - **After 5.0 Seconds:** \( 14.4 \, \text{m/s} \) downward