Question

A projectile is fired straight upward with an initial veloc- ity of 100 m=s from the top of a building 20 m high and falls to the ground at the base of the building. Find (a) its maximum height above the ground; (b) when it passes the top of the building; (c) its total time in the air.

Answer 1

In this problem, we calculate the maximum height, when it passes the top of the building, and the total time in the air for a projectile fired straight upward with an initial velocity of 100 m/s from the top of a building.

Step-by-step explanation:

In this problem, we are given a projectile fired straight upward from the top of a building with an initial velocity of 100 m/s. The projectile falls to the ground at the base of the building.

(a) To find the maximum height above the ground, we need to determine the time it takes for the projectile to reach its maximum height and then use that time to find the height. The formula to find the time of flight is:

time = (final velocity - initial velocity) / acceleration

Answer 2

The answer to your question is:

a) h max = 529.7 m

b) t = 20.4 s

c) t = 20.6 s

Step-by-step explanation:

a) h max = -(vo)² / 2g

= 100² / 2(9.81)

= 10000 / 19.62

= 509.7 m

total height = 509.7 + 20 = 529.7 m

b)

h = gt² / 2

t = √ 2h / g

t = √ 2(509.7)/9.81

t = √ 103.91

t = 10.19 s

total time = 2 x t = 2 x 10.19 = 20.4 s

c)

h = vot + 1/2gt²

20 = 100t + 1/2(9.91) t²

4.9t² + 100 t -20 = 0 quadratic equation

t = 0.19 s

Total time = 0.19 + 20.4 = 20.6 s