A projectile is fired up into the air at a speed of 164 m/s at an angle of 37° relative to the horizontal. Ignore air drag.

A.) Determine the MAXIMUM HEIGHT (above where it was thrown) the projectile will reach.
m

B.) Determine how LONG the projectile will be in the air. Assume it lands at the same height it was launched.
s

C.)Assuming it lands at the same height it was launched at, determine how FAR the projectile will land from where it was launched.
m

D.) Assuming it lands at the same height it was launched at, determine how FAST the projectile will be going upon impact.
m/s

E.) Assuming it lands at the same height it was launched at, determine the ANGLE (relative to the horizontal) of impact.
°

To solve this problem, we can use the equations of projectile motion. Let's break it down step by step.

A.) To determine the maximum height reached by the projectile, we need to find the vertical component of its velocity at the highest point. We can use the initial velocity and the launch angle to calculate this.

The vertical component of the initial velocity is given by:
v_y = v * sin(θ)
where v is the initial velocity (164 m/s) and θ is the launch angle (37°).

Thus, v_y = 164 m/s * sin(37°).

The projectile reaches its maximum height when the vertical component of its velocity becomes zero. This happens at the highest point of the projectile's trajectory.

So, to find the maximum height, we need to solve for time (t) when v_y = 0.

Using the equation:
v_y = v_0y + at
where a is the acceleration due to gravity (-9.8 m/s^2) and t is the time, we can rewrite it as:
0 = v_y + (-9.8 m/s^2)t

Solving for t:
t = -v_y / (-9.8 m/s^2) = v_y / 9.8 m/s^2

Now, we can substitute the previously calculated value of v_y to find the maximum height (h_max):
h_max = v_y * t - 0.5 * g * t^2

Substituting the known values:
h_max = (164 m/s * sin(37°)) * (v_y / 9.8 m/s^2) - 0.5 * 9.8 m/s^2 * (v_y / 9.8 m/s^2)^2

Simplifying this equation will give us the maximum height in meters.

B.) To determine the total time the projectile is in the air, we can use the same equation we used to find t in part A. We just need to multiply t by 2 since the projectile will spend an equal amount of time going up and coming back down.

C.) To determine how far the projectile will land from where it was launched, we need to find the horizontal component of the projectile's velocity. This can be calculated using the initial velocity and the launch angle.

The horizontal component of the initial velocity is given by:
v_x = v * cos(θ)
where v is the initial velocity (164 m/s) and θ is the launch angle (37°).

To find the horizontal distance traveled (d), we can use the equation:
d = v_x * t
where t is the total time the projectile is in the air (which we calculated in part B).

D.) To find the final velocity of the projectile upon impact, we can use the equation:
v_f = sqrt(v_x^2 + v_y^2)
where v_x is the horizontal component of the velocity and v_y is the vertical component of the velocity.

E.) Finally, to determine the angle of impact (relative to the horizontal), we can use the equation:
θ_impact = arctan(v_y / v_x)

By applying these equations, we can find the answers to all the parts of the problem.

To solve these questions, we can use the equations of motion for projectile motion. The projectile has an initial velocity of 164 m/s at an angle of 37° relative to the horizontal. We'll consider the initial height, h_0, to be zero.

A.) Determine the maximum height (above where it was thrown) the projectile will reach.

To find the maximum height, we need to determine the vertical component of the initial velocity.

Vertical component of initial velocity (v_y) = v * sin(angle)
v_y = 164 m/s * sin(37°)
v_y = 98.2 m/s (rounded to one decimal place)

Next, we can use the formula for maximum height:

Maximum height (h_max) = (v_y^2) / (2 * g)
where g is the acceleration due to gravity (9.8 m/s^2).

h_max = (98.2 m/s)^2 / (2 * 9.8 m/s^2)
h_max = 1001 m (rounded to the nearest meter)

Therefore, the maximum height the projectile will reach is 1001 meters above the launch point.

B.) Determine how long the projectile will be in the air.

To find the time of flight, we can use the vertical component of initial velocity:

Time of flight (t) = (2 * v_y) / g
t = (2 * 98.2 m/s) / 9.8 m/s^2
t = 20 s

Therefore, the projectile will be in the air for 20 seconds.

C.) Assuming it lands at the same height it was launched at, determine how far the projectile will land from where it was launched.

To find the horizontal distance, we can use the horizontal component of initial velocity:

Horizontal component of initial velocity (v_x) = v * cos(angle)
v_x = 164 m/s * cos(37°)
v_x = 131.4 m/s (rounded to one decimal place)

Horizontal distance (d) = v_x * t
d = 131.4 m/s * 20 s
d = 2628 m (rounded to the nearest meter)

Therefore, the projectile will land approximately 2628 meters away from where it was launched.

D.) Assuming it lands at the same height it was launched at, determine how fast the projectile will be going upon impact.

The vertical velocity of the projectile upon impact is the same as the initial vertical velocity since it lands at the same height.

Vertical component of final velocity (v_yf) = v_y = 98.2 m/s

To find the horizontal component of final velocity, we use the fact that there are no horizontal forces acting on the projectile (assuming no air drag). Therefore, the horizontal velocity remains constant.

Horizontal component of final velocity (v_xf) = v_x = 131.4 m/s

The final velocity can be found using Pythagoras' theorem:

Final velocity (v_f) = sqrt((v_xf^2) + (v_yf^2))
v_f = sqrt((131.4 m/s)^2 + (98.2 m/s)^2)
v_f = 165.6 m/s (rounded to one decimal place)

Therefore, the projectile will be going approximately 165.6 m/s upon impact.

E.) Assuming it lands at the same height it was launched at, determine the angle (relative to the horizontal) of impact.

To find the angle of impact (theta_f), we can use the components of the final velocity:

tan(theta_f) = v_yf / v_xf
theta_f = atan(v_yf / v_xf)
theta_f = atan(98.2 m/s / 131.4 m/s)
theta_f = 37°

Therefore, the angle of impact, relative to the horizontal, is 37°.