a. To find the height h(t) of the ball t seconds after throwing it, we can use the kinematic equation for vertical motion:
h(t) = h₀ + v₀t + (1/2)at²
Where:
h(t) is the height of the ball at time t
h₀ is the initial height (40m in this case)
v₀ is the initial velocity (2 m/s in the downward direction)
a is the acceleration due to gravity (-1.62 m/s², considering it acts in the opposite direction of the initial velocity)
t is the time elapsed in seconds
Plugging in the values, we have:
h(t) = 40 + 2t - (1/2)(1.62)t²
b. To find when the ball hits the surface of the moon, we need to determine the time at which the height h(t) becomes zero. We can set the equation h(t) = 0 and solve for t:
0 = 40 + 2t - (1/2)(1.62)t²
Rearranging the equation and setting it equal to zero:
(1/2)(1.62)t² - 2t - 40 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
Where a = (1/2)(1.62), b = -2, and c = -40. Plugging in these values:
t = (-(-2) ± √((-2)² - 4(1/2)(1.62)(-40))) / (2(1/2)(1.62))
Simplifying further will give us the values for t when the ball hits the surface of the moon.