Answer:
a) No
b) the rock must have a minimum initial speed of 6.79m/s for it to reach the top of the building.
Explanation:
Given:
Height of the wall = 3.95m
Initial height = 1.60m
Initial speed = 5.00m/s
distance between the initial height and wall top = 3.95 - 1.60 = 2.35m
Using the formula;
v^2 = u^2 + 2as ....1
Where v = final velocity, u = initial velocity, a = acceleration, s = distance travelled
From equation 1
s = (v^2 - u^2)/2a ...2
Since the rock t moving up,
the acceleration = -g = -9.8m/s2
s = maximum height travelled
v = 0 (at maximum height velocity is zero)
Substituting into equation 2
s = (0 - 5^2)/(2×-9.8) = 1.28m
Therefore, the maximum height is 1.28 from his initial height Which is less than the 2.35m of the wall from his initial height. So the rock will not reach the top of the wall
b) Using equation 1:
u^2 = v^2 - 2as
v = 0
a = -9.8m/s
s = 2.35m. (distance between the initial height and wall top)
u^2 = 0 - 2(-9.8 × 2.35)
u^2 = 46.06
u = √46.06
u = 6.79m/s
Therefore, the rock must have a minimum initial speed of 6.79m/s
