An projectile is launched upward with a velocity of 256 feet per second from the top of a 35 foot structure. what is the maximum height attained by the projectile?

1052.64 feet

Step-by-step explanation:

Step 1:

Given,

The projectile is launched upward with a velocity of 256 feet per second

The height of the structure is 35 foot.

Step 2 :

The velocity becomes 0 at the highest point.

The equation of motion is v² = u² +2as

where

v represents the final velocity

u represents the initial velocity

a represents the acceleration due to gravity

s is the height

Step 3:

Here  we have

u = 256 feet/sec

v = 0

a = -32.2 feet/sec²

Substituting in the above equation, we have

0 = 256² + 2 *(-32.2) *s

=>  -64.4 s = -65536

= > s = 65536/64.4 = 1017.64 feet

Hence the projectile travels 1017.64 feet from the top of the structure. Since the height of the structure is given as 35 feet, the maximum height attained by the projectile is 1017.64 +35 = 1052.64 feet

answer:  

step-by-step explanation:

here x represents the height of the pole,

and, the height of the shadow of the pole is 12 feet.

also, the angle of elevation from the top of the pole is 40°,

therefore,

by the low of sine,

which is the required equation for finding the value of x,