Question

A projectile is shot from the ground at an angle of `pi/4` and a speed of `8 m/s`. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

Answer 1

The distance is `=3.65m`

Explanation:

Resolving in the vertical direction `uarr^+`

initial velocity is `u_y=vsintheta=8*sin(1/4pi)`

Acceleration is `a=-g`

At the maximum height, `v=0`

We apply the equation of motion

`v=u+at`

to calculate the time to reach the greatest height

`0=8sin(1/4pi)-g*t`

`t=8/g*sin(1/4pi)`

`=0.577s`

The greatest height is

`h=(8sin(1/4pi))^2/(2g)=1.63m`

Resolving in the horizontal direction `rarr^+`

To find the horizontal distance, we apply the equation of motion

`s=u_x*t`

`=8cos(1/4pi)*0.577`

`=3.27m`

The distance from the starting point is

`d=sqrt(h^2+s^2)`

`=sqrt(1.63^2+3.27^2)`

`=3.65m`

Answer 2

Please check the animations carefully. The answer is given at the end of the explanation.

Explanation:

`"Projectile motion is a special case of two-dimensional motion. "`
`"An object found at the moment of shooting is seen."`

`"if we ignore the air resistance and gravity,the object move to infinity."`
`"But that's not really the case. The object is attracted by the ground"` `" and placed on a trajectory."`
`"The object drops freely from point J to point H."`

`"We must divide the motion horizontally and vertically into two parts."`

• The blue vector shows the horizontal component of the initial velocity.

`"The " V_x " vector can be calculated using " v_x=vi*cos theta`

• The magnitude and direction of blue vector does not change.
• *The blue vector allows the object to move horizontally. *

• The green vector shows the vertical component of the initial velocity.

`"The " V_y " vector can be calculated using " v_y=vi*sin theta`

`V_y " at any time is "v_y=v_i*sin theta -g*t`

• The magnitude and direction of green vector change.
• *The blue vector allows the object to move vertically. *
• the magnitude of green vector at maximum height is zero.

`"How can I calculate the elapsed time from initial point to peak point ?"`

`v_y=0("at maximum height)"`
`0=v_i*sin theta-g*t`
`v_i*sin theta=g*t`
`t=(v_i*sin theta)/g`

`"How can I calculate the maximum height?"`

`h_m=(v_i^2*sin^2 theta)/(2*g)`

`"How can I calculate the elapsed time for traveled from initial point to end point ?"`

`t=2*(v_i*sin theta)/2`

`"How do we find the velocity of a projectile at any time over its trajectory?"`

`"Calculate "v_x=v_i*cos theta`
`"Calculate "v_y=v_i*sin theta-g*t`
`"Calculate "v=sqrt((v_x)^2+(v_y)^2)`

`"How can I calculate the location of object on trajectory?"`

`x=v_i*t*cos theta`
`y=v_i*t*sin theta-1/2 g t^2`

`"How can I calculate the maximum x-range?"`

`x_m=(v_i^2*sin(2 theta))/g`

`"answer to your question."`
`theta=pi/4`
`v_i=8 m*s^(-1)`

`t=(v_i*sin theta)/g=(8*0.707)/(9.81)=0.58 s`

`x=v_i*t*cos theta=8*0.58*0.707=3.28m`