A catapult gives a rock a vertical velocity of 122m/s.

a. Determine the time the rock is in the air.
b. Determine the maximum height the rock would reach.
c. As the rock leaves the catapult the rock also has a horizontal velocity of 34m/s. solve for the rocks range.
d. Solve for the resultant velocity that the rock would impact the ground with.

To answer these questions, we need to use the equations of motion under constant acceleration.

a. The vertical velocity given is the initial vertical velocity. As the rock is projected vertically, the final vertical velocity will be zero at the highest point. The acceleration due to gravity is constant and equal to -9.8 m/s².

Using the equation:
v = u + at

We have:
0 = 122 m/s - 9.8 m/s² * t

Solving for time (t):
9.8 m/s² * t = 122 m/s
t = 122 m/s / 9.8 m/s²
t ≈ 12.45 seconds

Therefore, the rock is in the air for approximately 12.45 seconds.

b. To determine the maximum height reached by the rock, we can use the equation of motion for vertical displacement:

s = ut + (1/2)at²

Since the final vertical velocity is zero at the highest point, we can write:

0 = 122 m/s - 9.8 m/s² * t_max

Solving for t_max:
t_max = 122 m/s / 9.8 m/s²
t_max ≈ 12.45 seconds

Now, substituting the value of t_max into the equation of motion for vertical displacement, we get:

s_max = (1/2)(-9.8 m/s²)(12.45 s)²
s_max ≈ 764.6 meters

Therefore, the maximum height reached by the rock is approximately 764.6 meters.

c. The horizontal velocity of the rock is unaffected by gravity. Therefore, the time it takes for the rock to reach the ground horizontally is the same as the time it spends in the air vertically, which is approximately 12.45 seconds.

To find the range of the rock, we can use the equation of motion for horizontal displacement:

s = ut

Substituting the horizontal velocity (u = 34 m/s) and the time (t = 12.45 s), we get:

s = 34 m/s * 12.45 s
s ≈ 423.3 meters

Therefore, the range of the rock is approximately 423.3 meters.

d. The resultant velocity that the rock would impact the ground with can be found using the Pythagorean theorem. Since the horizontal and vertical components of velocity are perpendicular to each other, we can find the resultant velocity (v_impact) using the equation:

v_impact = √(v_horizontal² + v_vertical²)

Substituting the given velocities, we have:

v_impact = √(34 m/s)² + (122 m/s)²
v_impact ≈ √1156 m²/s² + 14884 m²/s²
v_impact ≈ √16040 m²/s²
v_impact ≈ 126.8 m/s

Therefore, the resultant velocity that the rock would impact the ground with is approximately 126.8 m/s.