A projectile is shot straight up from the earth's surface at a speed of 11,000 km/hr. How high does it go? km? Taken from "Physics for Scientists and Engineers by Randall D. Knight 2nd Edition. Chapter 13 #34. There is an answer in the database already, but I do not understand it.

476.35 km

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 11000 km/hr

Final velocity (v) = 0 km/hr (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s²

Maximum height (h) = ?

Next, we shall convert 9.8 m/s² to km/hr². This is illustrated below:

1 m/s² = 12960 km/hr²

Therefore,

9.8 m/s² = 9.8 m/s² × 12960 km/hr² / 1 m/s²

9.8 m/s² = 127008 km/hr²

Thus, 9.8 m/s² is equivalent to 127008 km/h²

Finally, we shall determine the maximum height reached by the projectile.

This is illustrated below:

Initial velocity (u) = 11000 km/hr

Final velocity (v) = 0 km/hr (at maximum height)

Acceleration due to gravity (g) = 127008 km/hr²

Maximum height (h) = ?

v² = u² – 2gh (since the projectile is going against gravity)

0² = 11000² – (2 × 127008 × h)

0 = 121×10⁶ – 254016h

Collect like terms

0 – 121×10⁶ = – 254016h

– 121×10⁶ = – 254016h

Divide both side by – 254016

h = – 121×10⁶ / – 254016

h = 476.35 km

Thus, the maximum height reached by the projectile is 476.35 km

a, c, d, e are all applying about success strategies.

the total distance traveled by the car is

30km + 49km = 79 km .