Ek = (m*V^2) / 2 where m is mass and V is speed, then we can take this equation and manipulate it a little to isolate the speed.
Ek = mv^2 / 2 — multiply both sides by 2
2Ek = mv^2 — divide both sides by m
2Ek / m = V^2 — switch sides
V^2 = 2Ek / m — plug in values
V^2 = 2*30J / 34kg
V^2 = 60J/34kg
V^2 = 1.76 m/s — sqrt of both sides
V = sqrt(1.76)
V = 1.32m/s (roughly)
Answer:
(a) 8 m/s
(b) 5 s
Explanation:
(a)
Using,
V² = U²+2gh ......................... Equation 1
Where V = final velocity, U = Initial velocity, g = acceleration due to gravity on the surface of the moon, h = height reached.
Given: V = 0 m/s ( At it's maximum height), g = -1.6 m/s² ( as its moves against gravity), h = 20 m.
Substitute into equation 1
0 = U²+[2×20×(-1.6)]
-U² = - 64
U² = 64
U = √64
U = 8 m/s.
(b)
V = U +gt.................... Equation 2
Where t = time to reach the maximum height.
Given: V = 0 m/s ( At the maximum height), g = -1.6 m/s² ( Moving against gravity), U = 8 m/s.
Substitute into equation 2
0 = 8+(-1.6t)
-8 = -1.6t
-1.6t = -8
t = -8/-1.6
t = 5 s.
