Agolfer hits a golf ball at an angle of 25.0â° to the ground. if the golf ball covers a horizontal distance of 301.5 m, what is the ballâs maximum height? (hint: at the top of its flight, the ballâs vertical velocity component will be zero.)

 Maximum height reached = 35.15 meter.

Explanation:

Projectile motion has two types of motion Horizontal and Vertical motion.

Vertical motion:

         We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

         Considering upward vertical motion of projectile.

         In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g and final velocity = 0 m/s.

        0 = u sin θ - gt

         t = u sin θ/g

    Total time for vertical motion is two times time taken for upward vertical motion of projectile.

    So total travel time of projectile = 2u sin θ/g

Horizontal motion:

  We have equation of motion , , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

  In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 and time taken = 2u sin θ /g

 So range of projectile,  

 Vertical motion (Maximum height reached, H) :

     We have equation of motion, , where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.

   Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H

In the give problem we have R = 301.5 m,  θ = 25° we need to find H.

So  

Now we have

 So maximum height reached = 35.15 meter.

speed

ratio of distance traveled to time interval

net force

vector sum of forces on object

object

source of diverging light rays either luminous or illuminated

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