An arrow is launched upward with a velocity of
320 feet per second from the top of a
50-foot platform. what is the matimum height attained by the arrow?

Step-by-step explanation:

The velocity of the arrow when it was launched upwards is 320 feet per second. The height of the platform from which it was launched is 50 ft

This means that the arrow is already 50 feets from the ground before moving upwards.

The height that it attains on moving from 50 feets will be determined by Newton's formula of motion.

v^2 = u^2 + 2as

u = initial velocity = 320 feet per second.

1 feet per second = 0.305 meters per second

320 feet per second = 320× 0.305

= 97.6 feets per second

a = acceleration due to gravity =

- 32.17405 feet/second^2 because the arrow is moving upwards)

s =maximum height of the arrow

v = final velocity = 0 feet/second

Substituting into the equation,

0^2 = 97.6^2 + 2×- 32.17405 × s

0 = 9525.76 - 64.3481s

64.3481s = 9525.76

s = 9525.76/64.3481 = 148.03 feets

Total height from the ground would be

148.03 + 50 = 198.03 feets

3; 9 is 8 to -4

step-by-step explanation: