Two boys are throwing a baseball back and forth. the ball is 4 ft above the ground when it leaves one child’s hand with an upward velocity of 36 ft/s. if acceleration due to gravity is –16 ft/s2, how high above the ground is the ball 2 s after it is thrown?

Step-by-step explanation:

It is given that, two boys are throwing a baseball back and forth.

Height above ground, h₀ = 4 ft

Initial velocity of child, u = 36 ft/s

Acceleration due to gravity, a = -16 ft/s²

The equation of projectile motion is given by :

According to the given condition,  

h(t) = 2.356 meters

Height of the ball at t = 2 s

h = 12 ft

So, at t = 2 seconds the height of the ball is 12 feet from the ground. Hence, this is the required solution.

the terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly. a recursive formula for a geometric sequence with common ratio r is given by an=ran−1 a n = r a n − 1 for n≥2 n ≥ 2 .

step-by-step explanation: