Answer:
2.583 s, 29.77 ft and 1.219 s
Explanation:
Using equation of motion and taken the motion upward as positive, also a = g ( acceleration due to gravity) = - 32 fts⁻², V= 39 fts⁻¹ V₁ is final velocity, y is the distance in ft from the ground
H = 6 ft, the height from which it is tossed
V₁ = V + gt = V - gt
at maximum height the body came to rest momentarily V₁ = 0
0 = V - gt
-V = -gt
- 39 / -32 = t
t time to reach maximum height = 1.219 s
To Maximum height reached can be calculated with the formula
V₁² = V² + 2g( y - H) where H is the initial height reached by the tossed rock
where V₁ is the final velocity at maximum height which = 0
0 = V² - 2g(y-H) where y is the distance traveled from the ground
-V² = -2g(y-H)
₋V² / -2g = y-H
(V²/2g) + H = y in ft
(39² / (2 × 32)) + 6
y = 29.77 ft
The total time it will be in air can be calculated with the formula below
y = H + Vt - 0.5gt² from y-H = ut + 0.5at²
0.5gt² - Vt - H = 0 since the body returned to the ground ( y = 0)
0.5gt² - Vt - H = 0
using quadratic formula
- (-V)² ± √ ((-V²) - 4 × 0.5g × -H) / (2 × 0.5 × g)
(V ± √ (V² + 2gH)) ÷ g
substitute the values into the expression
t = (39 + √(39² + (2×-32× 6)))/ 32 or (39 - √ (39² + (2 × -32×6))/ 32
t = (39 + √(1521 +384))/32 = (39 + √1905) / 32 = 2.583 s
t = (39 - √1905) / 32 = -0.15 s
The will remain in air (V ± √ (V² + 2gH)) / g seconds. It will reach a maximum height of (V²/2g) + H feet after V/g seconds
