Answer:
The ball will take 1 sec
Step-by-step explanation:
∵ The distance from the ground when the ball leaves = 6 ft
∵ The distance from the ground when the ball catches = 10 ft
∴ The difference between two position = 10 - 6 = 4 ft
∵ s = ut - 1/2 at² ⇒ s is the distance , u is the initial velocity , a is acceleration of gravity and t is the time
∵ s = 4 feet , u = 20 feet/sec and a ≅ 32 feet/sec²
∴ 4 = 20t -32/2 t²
∴ 16t² - 20t + 4 = 0
∴ (4t - 4)(4t - 1) = 0
∴ 4t - 4 = 0 ⇒ t = 1 and 4t - 1 = 0 ⇒ t = 1/4
∴ The ball will take 1 sec to go from first player to the second player
Note: 1/4 sec is the time of the ball after leaves the position of the first player by 4 feet above the initial position.
There’s no picture or reference
Answer:
Step-by-step explanation:
The answer is 20q^17w^9
Multiply 65 by .45 which is 29.25, and you add it to 65 since it says “markup” in the problem. So the answer is $94.25
60 divided by 4 is 15 15 multiplied by 3 is 45 pages
