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A brick is thrown upward from the top of a building at an angle of 25 degrees to the horizontal with an initial speed of 15 m/s. If the brick is in flight for 4 s before it hits the ground, how tall is the building?

Respuesta :

s1090322

Answer:

First, find the maximum height, which according to the values given, can be stated as:

H=(u²sin²theta)/2g

u=15m/s, theta=25 degrees, g=9.8m/s²

H= (15² * (sin 25)²))/2*9.8

H= (225*0.179)/19.6

H= 40.275/19.6

=2.06m

To find the velocity at maximum height:

Use the formula

v²=u²-2gH

It's minus because the brick was thrown upwards

So plugging everything into the above formula:

v²=15²-2*9.8*2.06

v²=225–40.376

v²=184.624

v=√184.624

v=13.59m/s

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