physics problem...A rock is thrown vertically upward...?
A rock is thrown vertically upward with a speed of 18.0m/s from the roof of a building that is 70.0m above the ground. Assume free fall.
In how many seconds after being thrown does the rock strike the ground?
What is the speed of the rock just before it strikes the ground?
Comments
First find the max height of the rock:
ymax = (V^2 - V0^2)/(2g) where v0 = 18 m/s V = 0 and g = -9.8 m/s^2
ymax = 18^2/(2*9.8) = 16.53 m from top of building.
Now rock falls 16.53 + 70 = 86.53 m and starts with V0 = 0 so
y = -86.53 = V^2/(2*g) ---> V =sqrt(2*9.8*86.53) = -41.18 m/s
Now we can compute time:
Time to go to max height is found using
v = gt+v0 = 0 =-9.8 t + 18 ---> t = 18/9.8 = 1.84 sec
Now time to go from max height to ground is:
v =gt = -41.18 = -9.8 t ---> t =41.18/9.8 = 4.2 sec
Total time = 4.2 + 1.84 = 6.04 sec
enable’s paintings backwards! through fact the rock falls from its maximum top, its speed will strengthen 9.8 m/s each and every 2d. on the utmost top the fee replaced into 0 m/s. enable’s confirm the time that the ball replaced into falling. vf = vi + 9.8 * t 4.9 = 0 + 9.8 * t Time down = 4.9 ÷ 9.8 = 0.5 2d The ball were moving for 2.5 seconds previously the fee replaced into 4.9 m/s down. 2.5 – 0.5 = 2.0 The ball were increasing for 2 seconds. through fact the ball replaced into increasing, the fee decreased 9.8 m/s each and every 2d. while the ball reached the utmost top, the fee replaced into 0 m/s. vf = vi – 9.8 * t vf = 0 m/s 0 = vi – 9.8 *2 vi = 19.6 m/s Vertical place = vi * t – ½ * 9.8 * t^2 t = 2.5 seconds Vertical place = 19.6 * 2.5 – ½ * 9.8 * 2.5^2 = 18.375 m the lady is eighteen.375 meters above the ingredient from which the rock replaced into thrown. OR enable’s confirm the utmost top and the area from the utmost top to the lady. The ball moved upward from its preliminary top to the utmost top in 2 seconds. So, the ball moved downward from the utmost top to the preliminary top in 2 seconds. on the utmost top the vertical speed is often 0 m/s. maximum top = Distance down d = vi * t + ½ * a * t^2 = ½ * 9.8 * 2^2 = 19.6 m the utmost top is nineteen.6 meters The balls from the utmost top for 0.5 seconds d = ½ * 9.8 *0.5^2 = a million.225 m The ball replaced right into a million.225 meters decrease than the utmost top while the lady observed it. 19.6 = a million.225 = 18.375 m the lady is eighteen.375 meters above the ingredient from which the rock replaced into thrown. making use of two techniques permits you to examine your answer!