Answer:
Acceleration = (change in speed) / (time for the change)
Change in speed= (0 - 26 km/hr) = -26 km/hr
(-26 km/hr) x (1,000 m/km) x (1 hr / 3,600 sec) = -7.222 m/sec
Average acceleration = (-7.222 m/s) / (22 min x 60sec/min) = -0.00547 m/sec²
Average speed during the stopping maneuver =
(1/2) (start speed + end speed) = 13 km/hr = 3.6111 m/sec
Explanation:
Answer:
The total energy, i.e. sum of kinetic and potential energy, is constant.
i.e. E = KE + PE
Initially, PE = 0 and KE = 1/2 mv^2
At maximum height, velocity=0, thus, KE = 0 and PE = mgh
Since, total energy is constant (KE converts to PE when the ball is rising),
therefore, KE = PE
or, 1/2 mv^2 = mgh
or, h = v^2 /2g = 13^2 / (2x9.8) = 8.622 m
Hope this helps.
