Answer:
0.131 m
Explanation:
Applying,
v² = u²+2gh................... Equation 1
Where, v = final velocity, u = initial velocity, g = acceleration due to gravity, h = maximum height of the object
make h the subject of the equation
h = (v²-u²)/2g........... Equation 2
From the question,
Given: v = 0 m/s (at the maximum height), u = 1.60 m/s
Constant: g = -9.8 m/s² (thrown up)
Substitute these values into equation 2
h = (0²-1.6²)/[2×(-9.8)]
h = -2.56/-19.6
h = 0.131 m
