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

h = 21.5 m

Step-by-step 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 -(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 = (\Delta x)/(v_(ox) ) = (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 -(1)/(2)*(9.8m/s2)*(3.75s)^(2) = -21.5 m (5)`

• ⇒ h = -(-21.5m) = 21.5 m