1.) A ball is thrown straight upwards at a speed of 32.5 m/s and from a height of 1 m above the ground. How long before it hits the ground?

2.) An astronaut jumps on the moon with an initial upward velocity of 4.35m/s. If the acceleration of gravity on the moon is -1.62m/s^2, what is the max height of the astronaut?

1.) To find the time it takes for the ball to hit the ground, we can use the kinematic equation:

v = u + at

where:
v = final velocity (0 m/s when the ball hits the ground)
u = initial velocity (32.5 m/s)
a = acceleration (acceleration due to gravity, -9.8 m/s^2)
t = time

Rearranging the equation, we have:

t = (v - u) / a

Plugging in the values, we get:

t = (0 - 32.5) / -9.8

Simplifying the equation, we have:

t = 3.316 seconds

Therefore, it will take approximately 3.316 seconds for the ball to hit the ground.

2.) To find the maximum height of the astronaut on the moon, we can use the kinematic equation:

v^2 = u^2 + 2as

where:
v = final velocity (0 m/s when the astronaut reaches the maximum height)
u = initial velocity (4.35 m/s)
a = acceleration (acceleration due to gravity on the moon, -1.62 m/s^2)
s = displacement (maximum height)

Re-arranging the equation, we have:

s = (v^2 - u^2) / (2a)

Plugging in the values, we get:

s = (0^2 - 4.35^2) / (2 * -1.62)

Simplifying the equation, we have:

s = 5.318 meters

Therefore, the maximum height the astronaut reaches on the moon is approximately 5.318 meters.

1.) To find the time it takes for the ball to hit the ground, we can use the equation of motion for vertical motion:

\(h = h_0 + v_0t + \frac{1}{2}gt^2\)

Where:
\(h\) is the final height (in this case, 0 since it hits the ground)
\(h_0\) is the initial height (1m in this case)
\(v_0\) is the initial velocity (32.5 m/s in this case)
\(g\) is the acceleration due to gravity (approximately -9.8 m/s^2)

Now, let's plug the values into the equation and solve for \(t\):

\(0 = 1 + 32.5t + \frac{1}{2}(-9.8)t^2\)

Simplifying the equation:

\(-4.9t^2 + 32.5t + 1 = 0\)

This is a quadratic equation, and we can solve it using various methods such as factoring, completing the square, or using the quadratic formula. In this case, let's use the quadratic formula:

\(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Plugging in the values:

\(a = -4.9\)
\(b = 32.5\)
\(c = 1\)

\(t = \frac{-32.5 \pm \sqrt{(32.5)^2 - 4(-4.9)(1)}}{2(-4.9)}\)

Now, we can calculate the roots:

\(t_1 \approx 0.0808 \, \text{s}\) (ignoring the negative solution since time cannot be negative)

Therefore, it takes approximately 0.0808 seconds for the ball to hit the ground.

2.) To find the maximum height of the astronaut, we need to determine the time it takes for the astronaut to reach the highest point of their jump. At this point, the vertical velocity will be zero because the astronaut will momentarily stop moving upwards before falling back down.

We can use the equation of motion:

\(v = v_0 + gt\)

Where:
\(v\) is the final velocity (0 m/s in this case)
\(v_0\) is the initial velocity (4.35 m/s in this case)
\(g\) is the acceleration due to gravity on the moon (-1.62 m/s^2 in this case)
\(t\) is the time we're trying to find

Substituting the known values:

\(0 = 4.35 - 1.62t\)

Now, let's solve for \(t\):

\(1.62t = 4.35\)

\(t \approx 2.684 \, \text{s}\)

The time taken for the astronaut to reach the maximum height is approximately 2.684 seconds.

To find the maximum height, we can use the equation of motion for vertical motion:

\(h = h_0 + v_0t + \frac{1}{2}gt^2\)

At the maximum height, the final velocity is zero, so:

\(h = h_0 + v_0t + \frac{1}{2}gt^2\)

Plugging in the values:

\(h = 0 + 4.35(2.684) + \frac{1}{2}(-1.62)(2.684)^2\)

Simplifying the equation:

\(h \approx 2.93 \, \text{m}\)

Therefore, the maximum height reached by the astronaut is approximately 2.93 meters.