An arrow is shot from a height of 1.5 m toward a cliff of height H. It is shot with a velocity of 30 m/s at an angle of 60 degrees above the horizontal. It lands on the top edge of the cliff 4.0 s later. (a) What is the height of the cliff

The height of the cliff is 27.02 m

Explanation:

Given;

height above the ground from which the arrow is shot, h₁ = 1.5 m

initial velocity of the projectile, v = 30 m/s

angle of projection, θ = 60⁰

time taken to reach top of the cliff, t = 4.0 s

The vertical component of the velocity is calculated as;

The height attained by the projectile at the given time is calculated as;

The height of the cliff is calculated as;

H = h₁ + h₂

H = 1.5 m  +  25.52 m

H = 27.02 m

are there options? cause you ain’t. real specific

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s