Answer:
a. The football with initial vertical velocity 44 ft per second b. Time in air for 44 ft per second ball = 2.76 s . Time in air for 40 ft per second ball = 2.50 s
Explanation:
a. The football with initial vertical velocity 44 ft per second
b. Using v = u + at where v = velocity at maximum height = 0
For first football u = 44 ft/s,a = g = -32 ft/s²
v = u + at
0 = 44 + (-32)t
0 = 44 -32t
-44 = -32t
t ₁= 44/32 = 1.38 s
The vertical distance it moves is gotten from v² = u² + 2as with v = 0,
s = u²/2a = 44²/(2 × 32) = 30.25 ft
Since it covers this same distance on its downward fall, its velocity as it as it hits the ground is v² = u² + 2as where u = 0 and g = -32ft/s²
v² = u² + 2as = 0 + 2 ×(-32) ×(-30.25) = 1936 ⇒ v =√1936 = 44 ft/s
The time it takes to cover this distance is gotten from v = u + at with u = 0
-44 = 0 + (-32)t
-44 = 0 -32t
-44 = -32t
t₂ = 44/32 = 1.38 s
total time = t₁ + t₂ = 1.38 s + 1.38 s = 2.76 s
For second football u = 40 ft/s,a = g = -32 ft/s²
v = u + at
0 = 40 + (-32)t
0 = 40 -32t
-40 = -32t
t₃ = 40/32 = 1.25 s
The vertical distance it moves is gotten from v² = u² + 2as with v = 0,
s = u²/2a = 40²/(2 × 32) = 25 ft
Since it covers this same distance on its downward fall, its velocity as it as it hits the ground is v² = u² + 2as where u = 0 and g = -32ft/s²
v² = u² + 2as = 0 + 2 ×(-32) ×(-25) = 1600 ⇒ v =√1600 = 40 ft/s
The time it takes to cover this distance is gotten from v = u + at with u = 0
-40 = 0 + (-32)t
-40 = 0 -32t
-40 = -32t
t₄ = 40/32 = 1.25 s
total time = t₃ + t₄ = 1.25 s + 1.25 s = 2.50 s
