Please help me with both problems I’ll mark as brainliest!!!
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Answer:
The baseball reached 9 feet high above the ground
Step-by-step explanation:
The given quadratic function representing the position of the baseball above ground is p(t) = 1/2·g·t² + v₀·t + p₀
Where;
t ≥ 0
g = -32 ft./sec²
v₀ = The initial velocity
p₀ = The initial position
Given that when the ball is thrown, we have;
The initial, straight up, velocity, v₀ = 16 ft./sec
The initial position, p₀ = 5 ft.
Substituting the above values in the quadratic function representing the position of the baseball above ground, we have;
p(t) = 1/2·(-32)·t² + 16·t + 5 = 16·t - 16·t² + 5
At the maximum point, the rate of change of the height with time = 0, therefore;
dp(t)/dt = 0 = d(16·t - 16·t² + 5)/dt = 16 - 32·t
16 - 32·t = 0
16 = 32·t
t = 16/32 = 0.5 seconds
Therefore, the time takes to reach the maximum height = 0.5 seconds
The height (maximum) reached in 0.5 seconds is given as follows;
h(t) = 16·t - 16·t² + 5, from which we have;
h(0.5) = 16 × 0.5 - 16 × (0.5)² + 5 = 9
Therefore, the height baseball reached = 9 ft. above ground
