PLEASE HELP An object is launched horizontally from a cliff. The cliff is 80 m high and the object has an initial launch velocity of 50 m/s.

What is the initial horizontal velocity?
What is the initial vertical velocity?
What is the final horizontal velocity?
How much time did it take for the object to hit the ground?
What is the final vertical velocity?
What is the final resultant speed?
How far from the base of the cliff will the projectile land?

a. 50 m/s b. 0 m/s c. 50 m/s d. 4.04 s e. -39.6 m/s f. 63.78 m/s g. 202 m

Explanation:

a. What is the initial horizontal velocity?

Since the object is launched horizontally, it initial horizontal velocity is 50 m/s

b. What is the initial vertical velocity?

Since the object is launched horizontally, it has no initial vertical component. So, its initial vertical velocity is 0 m/s

c. What is the final horizontal velocity?

Its final horizontal velocity is 50 m/s since no force acts on it in the horizontal direction to change its value.

d. How much time did it take for the object to hit the ground?

We use the equation s = ut - 1/gt² since the object is falling under gravity where u = initial vertical velocity = 0 m/s, s = height of cliff = 80 m, g = acceleration due to gravity = -9.8 m/s² and t = time it takes the object to hit the ground.

s = ut - 1/2gt²

80 m = 0 × t - 1/2 × -9.8 m/s² × t²

80 m = 4.9 m/s² × t²

t² = 80 m ÷ 4.9 m/s²

t² = 16.33 s²

t = √(16.33 s²)

t = 4.04 s

e. What is the final vertical velocity?

Using v = u + at where u = initial vertical velocity = 0 m/s, v = final vertical velocity, a = acceleration = -g =  -9.8 m/s²and t = time it takes object to reach the ground = 4.04 s.

Substituting these values into the equation, we have

v = u + at

v = 0 m/s + (-9.8 m/s²) × 4.04 s

v = -39.6 m/s

f. What is the final resultant speed?

The final resultant speed v' is the resultant of the final horizontal velocity and the final vertical velocity. Let u' = final vertical velocity = 50 m/s.

v' = √(u'² + v²)

v' = √((50 m/s)² + (-39.6 m/s)²)

v' = √(2500 m²/s² + 1568.16 m²/s²)

v' = √(4068.16 m²/s²)

v' = 63.78 m/s

g. How far from the base of the cliff will the projectile land?

The distance from the base of the cliff, d where the projectile lands is

d = u't where u' = horizontal velocity = 50 m/s and t = time it takes object to land = 4.04 s

d = 50 m/s × 4.04 s

d = 202 m

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.