Question

g You are walking around your neighborhood and you see a child on top of a roof of a building kick a soccer ball. The soccer ball is kicked at 37° from the edge of the building with an initial velocity of 21 m/s and lands 63 meters away from the wall. How tall, in meters, is the building that the child is standing on?

Answer 1

Answer:

h = 21.5 m

Explanation:

• First of all, we define a pair of coordinate axes along the horizontal and vertical direction, calling x-axis to the horizontal and y-axis to the vertical, with the origin in the point where the ball is kicked.
• Neglecting air resistance, the only influence on the ball once kicked is due to gravity, so the ball is accelerated by the Earth with a constant value of -9.8 m/s2 (assuming the upward direction as positive).
• So, we can use the kinematic equation for displacement for the vertical direction, as follows:

       $\Delta y = v_{oy}* t -\frac{1}{2}*g*t^{2} (1)$

• Since the ball is kicked at an angle of 37º from the edge of the building, at an initial velocity of 21 m/s, we can find the horizontal and vertical initial speeds as follows:

       $v_{ox} = v* cos 37 = 21 m/s * cos 37 = 16.8 m/s (2)$

       $v_{oy} = v* sin 37 = 21 m/s * sin 37 = 12.6 m/s (3)$

• In the horizontal direction, since gravity has no component in this direction, the ball moves at a constant speed, equal to v₀ₓ.
• Applying the definition of average velocity, since we know the horizontal distance traveled, we can find the total time that the ball was in the air, as follows:

       $t = \frac{\Delta x}{v_{ox} } = \frac{63m}{16.8m/s} = 3.75 s (4)$

• Replacing (4) and (3) in (1), we can find the total vertical displacement, which is equal to the height of the building, as follows:

     $-h = 12.6m/s* 3.75s -\frac{1}{2}*(9.8m/s2)*(3.75s)^{2} = -21.5 m (5)$

• ⇒ h = -(-21.5m) = 21.5 m