Calculate the initial velocity of the projectile

Help with physics problem and solution

A ball is launched from the ground at some angle to the horizontal ground. At a height h₁=6m the components of velocity are: v_x=5.0m/s, v_y=4.0m/s. (see figure).

(1) What was the magnitude of the initial velocity v₀? (2) what was the direction of the initial velocity with respect to ground level? (3) What was the maximum height H of this projectile?

Given are:

h₁=6m,

v_x=5.0m/s,

 v_y=4.0m/s.

We are looking for:

v₀=? - magnitude of initial velocity,

α = ? – direction of v₀ with respect to ground level,

H = ? – the maximum height of this projectile

Solution:

From the paragraph “Projectile motion” we know that the horizontal component of velocity is constant during projectile motion, so we can write

v_ox = v_o           (1)

From the same paragraph we know that the height at time t₁ during the projectile motion is given by

h₁ = v_0y t₁ - (1/2)g t₁²    (2)  

The vertical component of velocity is given by formula

v_y = v_oy – gt₁          (3)

 

Equations (2) and (3) constitute a set of 2 equations with 2 unknowns – v_0y and t₁. This is not a set of linear equations. It can be solved by different methods, here we present one of them.

We solve (2) with respect to time t₁.

t₁ = (v_0y – v_y)/g       (4)

Now t₁ from equation (4) we substitute into equation (2) and with help of algebra we can write.

       (5)

         (6)

          (7)

     (8)

            (9)

 

So

     (10)

We check dimensions

[v_0y] = {(m/s²)m}^-(1/2) = m/s

as the velocity dimension should be. Substituting numbers given in the problem we get

v_0y = 11.56m/s

The magnitude of initial velocity can be calculated (from the Pitagoras formula) as

     (11)

Substituting numbers given we get

v₀ = 12.59m/s  

and this is the answer to question (1).

The direction of this velocity with respect to ground level can be calculated from the definition of the tangent function. From figure in this problem we have

tanα = v_0y/v_0x     (12)

so

α = tan^-1(v_0y/v_0x)   (13)

α = 38.7^o            (14)

and this is the answer to question (2).

And now question (3).

With the help of paragraph “Projectile motion” we can answer this. The maximum height of a projectile is given by equation

    (15)

 For launching at angle 90º the v0 can be replaced by v0y and with sin90 º =1 the formula (15) becomes

   (16)

Substituting equation (10) into (16) we get

 

     (17)

which is the answer to question (3). Substituting numbers and with help of a calculator we  get the maximum height of this projectile

H = 6.82m

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