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?

2 Answers
May 29, 2017

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

May 29, 2017

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."

enter image source here

"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."

enter image source here

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

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

enter image source here

"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. *

enter image source here

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

enter image source here

"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

enter image source here

  • 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?"

enter image source here

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?"

enter image source here

"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?"

enter image source here

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