An object is projected from a height of 100m above the ground at an angle of 300to the horizontal with a velocity of 100m/s.
Calculate
(4)
(4)
The maximum height reached above the ground
Time of flight
The velocity and the direction of the object 1 sec before it hit the ground
(4)​

a) y = 127.6 m, b) 11.9s,  c)  v = 103.6  m / s, θ’= 326.7º

Explanation:

This is a missile throwing exercise

let's start by breaking down the initial velocity

         sin 30 =

         cos 30 = v₀ₓ / v₀

         v_{oy} = v₀ go sin 30

         v₀ₓ = v₀ cos 30

         v_{oy} = 100 sin 30 = 50 m / s

         v₀ₓ = 100 cos 30 = 86.6 m / s

a) the maximum height is requested.

At this point the vertical velocity is zero (v_y = 0)

         v_y² = - 2 g y

         0 = v_{oy}^2 - 2g y

         y =

         y = 50² / (2  9.8)

         y = 127.6 m

b) Flight time

this is the time it takes to reach the ground, the reference system for this movement is taken on the ground this is a height of y = 0 m and the body is at an initial height of i = 100m

         y = y₀ + v₀ t - ½ g t²

       0 = 100 + 50 t - ½ 9.8 t²

we solve the quadratic equation

        4.9 t² - 50 t - 100 = 0

         t =

         t =

         t₁ = 11.9 s

         t₂ = -8.4 s

flight time is 11.9s

c) The time 1 s before hitting the ground is

        t1 = 11.9 -1

        t1 = 10.9 s

let's find the vertical speed

       v_y = - g t

       v_y = 50 - 9.8 10.9

       v_y = -56.8 m / s

the negative sign indicates that the direction of the velocity is downward.

On the x-axis there is no acceleration therefore the speed is constant.

Let's use the Pythagorean theorem

        v =

        v =

        v = 103.6  m / s

let's use trigonometry

        tan θ =

        θ = tan⁻¹  \frac{v_y}{v_x}

        θ = tan⁻¹  (-56.8 / 86.60)

        θ = -33.3º

the negative sign indicates that it is measured clockwise from the x-axis

for a counterclockwise measurement

          θ’= 360 - θ

          θ' = 360 - 33.3

          θ’= 326.7º

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.