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 calculate the speed of the rock at launch, we can use the fact that the rock is under constant acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \) downward). Given that the speed 2.0 seconds after launch is \( 15 \, \text{m/s} \) upward, we can use the following kinematic equation: \[ v = u + at \] where: - \( v = 15 \, \text{m/s} \) (final speed after 2.0 s), - \( u \) is the initial speed (what we want to find), - \( a = -9.81 \, \text{m/s}^2 \) (negative because it’s acting against the motion), - \( t = 2.0 \, \text{s} \). Rearranging the equation to find \( u \): \[ u = v - at = 15 \, \text{m/s} - (-9.81 \, \text{m/s}^2)(2.0 \, \text{s}) = 15 \, \text{m/s} + 19.62 \, \text{m/s} = 34.62 \, \text{m/s} \] So, the speed at launch (just before the blast) is approximately \( 34.62 \, \text{m/s} \). Now for the speed 5.0 seconds after launch, we will again use the first formula: \[ v = u + at \] Using \( u = 34.62 \, \text{m/s} \): \[ v = 34.62 \, \text{m/s} + (-9.81 \, \text{m/s}^2)(5.0 \, \text{s}) = 34.62 \, \text{m/s} - 49.05 \, \text{m/s} = -14.43 \, \text{m/s} \] The negative sign indicates the rock is now falling downwards. Thus, 5.0 seconds after launch, the speed of the rock is approximately \( 14.43 \, \text{m/s} \) downward.