Question

Tom standing on the roof of a tall building of 60m height throws a ball at angle of 45⁰ .The velocity of the ball is 12 m/s when it leaves tom’s hand.

Calculate the following:

a-Initial velocity in X, and Y direction

b- Time to reach to its maximum height.

C-Maximum height of the ball with respect to the ground.

d- Final velocity –magnitude.

e- Velocity of the ball at maximum height.

f- Acceleration of the ball at maximum height.

g- Total time in the air.

h- Distance between the landing spot and the base of the building.

Answer 1

Answer:

a) $v_{y}$ = 8.485 m / s , vₓ = 8.485 m / s, b)   t = 0.866 s , c) y = 63.67 m , d) vₓ = 8.485 m / s , $v_{y}$ = -8.485 m / s, e)   v = vₓ = 8.485 m / s , f) $a_{y}$ = -g = -9.8 m / s², g)  t₁ = 4.93 s , h)  x = 41.83 m

Explanation:

This is a projectile launch exercise

a) let's use trigonometry to find the components of velocity

         sin 45 = $v_{y}$ / v

         cos 45 = vₓ / v

          $v_{y}$ = v sin 45

          vₓ = v cos 45

           $v_{y}$ = 12 sin 45

           vₓ = 12 cos 45

           $v_{y}$ = 8.485 m / s

           vₓ = 8.485 m / s

b) At the point of maximum height, the vertical speed (vy) is zero

          $v_{y}$ = $v_{oy}$ - g t

          t = $v_{oy}$ / g

          t = 8.485 /9.8

          t = 0.866 s

c) the vertical speed is zero

         $v_{y}$² = $v_{oy}$² - 2 g (y-y₀)

         y-y₀ = $v_{oy}$² / 2g

         y = 8.485² / (2 9.8) + 60

         y = 63.67 m

d) the final speed when it reaches mass height is the same output speed, but with the component and negative

            vₓ = 8.485 m / s

           $v_{y}$ = -8.485 m / s

e) At the point of maximum height the vertical speed is zero

            v = vₓ = 8.485 m / s

f) The acceleration of the ball depends on the Earth and vouchers

           $a_{y}$ = -g = -9.8 m / s²

On the x axis there is no acceleration

g) the total time this is to the point y = 0

          y-y₀ = $v_{o}$ t - ½ g t²

replace

          0-60 = 12 t - ½ 9.8 t²

          4.9 t² - 12 t - 60 = 0

           t² - 2.45 t - 12.25 = 0

We solve the second degree equation

          t = [2.45 + - √ 2.45² - 4 (-12.25)] / 2

          t = [2.45 + - √ (6.00 +49)] = (2.45 + - 7.416) / 2

We have two results.

         t₁ = 4.93 s

         t₂ = -2.48 s

Since time cannot be negative, the correct result is t₁ = 4.93 s

h) the horizontal distance is

          x = vₓ t

          x = 8.485 4.93

          x = 41.83 m