A brick is thrown upward from the top of a building at an angle of 250 above the horizontal, and with an initial speed of 15 m/s. If it takes 3.0 s for the brick hit the ground, (a) how tall is the building? (b) What is the maximum height of the brick?

(a) 25.08 m

(b) 2.05 m

Step-by-step explanation:

Let the height of the building be 'h'.

Given:

Angle of projection is, °

Initial speed is,

Time of flight is,

(a)

Consider the vertical motion of the brick.

Vertical component of initial velocity is given as:

Vertical displacement of the brick is equal to the height of the building.

So, vertical displacement = (Negative sign implies downward motion)

Acceleration is due to gravity in the downward direction. So,

Acceleration is,

Now, using the following equation of motion;

Therefore, the building is 25.08 m tall.

(b)

Let the maximum height be 'H'.

At maximum height, the vertical component of velocity is 0 as the brick stops temporarily in the vertical direction.

So,

Now, using the following equation of motion, we have:

Therefore, the maximum height of the brick is 2.05 m.

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step-by-step explanation:

horizontal translation of 2 units in the positive x direction

followed by

vertical translation of 3 units in the positive y direction