A body is projected at an angle of 30degrees to the horizontal with a speed of 30m/s. What will be the angle with the horizontal after 1.5sec. Take g as 10m/s^2

Given Information:

Launch angle of projectile = 30°

Initial velocity = V₀ = 30 m/s

Acceleration due to gravity = g = 10 m/s²

Required Information:

Angle with the horizontal after 1.5 sec = ?

The angle of the projectile to the horizontal after t = 1.5 seconds is 0°

Step-by-step explanation:

The horizontal component of the velocity is given by

Where V₀ is the initial velocity and θ is the launch angle

The vertical component of the velocity is given by

Where V₀ is the initial velocity, θ is the launch angle, g is the acceleration due to gravity and t is the time.

So after t = 1.5 sec

The horizontal component of the velocity is

And the vertical component of the velocity is

The angel is

Therefore, the angle of the projectile to the horizontal after t = 1.5 seconds is 0°

d

step-by-step explanation:

by observation, we can see that the the graph is simply the graph of y = |x| that has been shifted 4 units downwards and 3 units to the left.

we can recognize in choice d :

y = | x + 3 |   - 4 (rewriting in general form)

y = | x - (- 3) |   - 4

the "-4 " term means   that the graph has been shifted by 4 units in the negative y direction

the "-3 " term means   that the graph has been shifted by 3 units in the negative x direction

this matches our observation of the graph described above.

the square root of what number? if you add that, then someone will probably answer the question.