A ball is thrown up into the air from a height of 1.2m. if it takes 3.4 s to land, what was the velocity of the throw, and what was the maximum height of the ball?

To find the velocity of the throw and the maximum height of the ball, we can use the basic equations of motion.

Let's start by finding the velocity of the throw. We'll use the equation of motion:

v = u + at

where:
v = final velocity (which is zero when the ball lands)
u = initial velocity (which we want to find)
a = acceleration due to gravity (approximately -9.8 m/s^2, taking into account the direction)

Since the ball is thrown upwards, we can assume the acceleration due to gravity is negative. Thus,

0 = u - 9.8 * 3.4

Simplifying the equation, we have:

0 = u - 33.32

Rearranging the equation, we find:

u = 33.32 m/s

The velocity of the throw is approximately 33.32 m/s.

Now, let's calculate the maximum height of the ball. We'll use the equation:

h = u * t + (1/2) * a * t^2

where:
h = maximum height of the ball (which we want to find)
u = initial velocity (33.32 m/s)
t = time taken to reach maximum height (half of the total time, 3.4 s / 2 = 1.7 s)
a = acceleration due to gravity (-9.8 m/s^2)

Plugging in the values, we have:

h = 33.32 * 1.7 + (1/2) * (-9.8) * (1.7^2)

Simplifying the equation, we get:

h = 56.644 - 14.0012

Thus,

h = 42.6428 m

The maximum height of the ball is approximately 42.6428 meters.