Question

A projectile is launched at an angle 60° to the horizontal

with some initial speed vi-45 m/s, and air resistance is
negligible. The cliff is 265 m high.
A) Find the final velocity as a vector,
B) Find the maximum height.
C) Find the horizontal range of the motion.

Answer 1

To find the final velocity, maximum height, and horizontal range of a projectile launched at an angle, we can use the equations of projectile motion.

Step-by-step explanation:

To answer the given questions, we can use the equations of projectile motion.

A) The final velocity of the projectile can be broken down into its horizontal and vertical components. The horizontal component of velocity remains constant throughout the motion, and the vertical component changes. The final velocity can be found by using the equations:

• Vx = (initial velocity)(cos(angle))
• Vy = (initial velocity)(sin(angle)) - (acceleration due to gravity)(time)
• Velocity = √(Vx² + Vy²)

B) To find the maximum height, we need to find the time taken for the projectile to reach its highest point. This can be done by setting the vertical component of velocity equal to zero and solving for time:

• Vy = (initial velocity)(sin(angle)) - (acceleration due to gravity)(time)
• Setting Vy = 0, we can solve for time.

C) The horizontal range of motion can be found by multiplying the horizontal component of velocity (Vx) by the time taken for the projectile to reach the ground.

Using the given values, we can calculate the final velocity, maximum height, and horizontal range of the projectile.

Answer 2

Step-by-step explanation:

We have,

A projectile is launched at an angle 60° to the horizontal with some initial speed 45 m/s.

The cliff is 265 m high.

(A) The final velocity of the projectile is given by :

`v_f^2-v_i^2=2gs\\\\v_f^2=2gs+v_i^2\\\\v_f^2=2* 9.8* 265+45^2\\\\v_f=84.96\ m/s`

(B) The maximum height of the projectile is given by :

`H=(v_i^2\sin^2\theta)/(2g)\\\\H=(45^2* \sin^2(60))/(2* 9.8)\\\\H=77.48\ m`

(C) The horizontal range of the motion is given by :

`R=(v_i^2\sin 2\theta)/(g)\\\\R=((45)^2* \sin 2(60))/(9.8)\\\\R=178.94\ m`