A brick is thrown horizontally off the top of a building at 7 m/s. If the building is 31m tall, how far from the base of the building will the brick land.
time for the brick to fall 31 m
31=1/2 9.8 t^2 in the vertical
t=sqrt(62/9.8)
how far horizontally?
d=7*t
Where is the 62 coming from?
2*31
To find the horizontal distance the brick will travel before landing, we need to first determine the time it takes for the brick to fall from the top of the building.
In this case, the brick is initially thrown horizontally with a velocity of 7 m/s. Since there is no vertical component to this initial velocity, the brick will only be affected by the force of gravity vertically. We can use the formula for calculating the time of flight for an object under free fall to find the time it takes for the brick to fall:
š” = ā(2ā/š)
where š” is the time of flight, ā is the height of the building, and š is the acceleration due to gravity (approximately 9.8 m/sĀ²).
Substituting the given values:
š” = ā(2 * 31 / 9.8)
š” ā ā(62/9.8)
š” ā ā6.3265
š” ā 2.51 s (rounded to two decimal places)
Now that we have the time it takes for the brick to fall, we can calculate the horizontal distance using the formula for horizontal displacement:
š = š£š„ * š”
where š is the horizontal distance, š£š„ is the horizontal velocity (which remains constant because there is no horizontal force affecting the brick's motion), and š” is the time.
Substituting the given values:
š = 7 * 2.51
š ā 17.57 m (rounded to two decimal places)
Therefore, the brick will land approximately 17.57 meters from the base of the building.