Question

An attacker at the base of a castle wall 3.60 m high throws a rock straight up with speed 8.00 m/s from a height of 1.70 m above the ground. Does the change in speed of the downward-moving rock agree with the magnitude of the speed change of the rock moving upward between the same elevations?

Answer 1

we can say here that | v² - u² | is the same for upward as for downward and change in the speed is different here so | v - u | same whenever rock travel up, down for same time and not same distances

Step-by-step explanation:

given data

base = 3.60 m

speed u = 8 m/s

height = 1.70 m

to find out

check change in speed

solution

we know here formula for v that is

v² = u² - 2gh ............1 for upward speed

v² = u² + 2gh ............2 for projected speed

so here put all value and find v with h = 3.60 - 1.70 = 1.9 m

v² = 8² - 2(9.8) 1.9 = 26.76

v² = 8² + 2(9.8) 1.9 = 101.24

v = 5.173 m/s ..............3

v = 10.061 m/s ...................4

so change in speed form 3 and 4 equation

change in speed = v - u = 8 - 5.173 = 2.827 m/s .................5

change in speed = v - u = 10.061 - 8 = 2.061 m/s ..................6

so now we can say here that | v² - u² | is the same for upward as for downward and change in the speed is different here so | v - u | same whenever rock travel up, down for same time and not same distances