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
I used bars of width 10. There are 3 values from 0-9; 4 values from 10-19; 3 values from 20-29; 0 values from 30-39; 6 values from 40-49; and 1 value from 50-59.
