Answer:
The height of the baseball is 35 feet at the moment the player begins to leap.
Step-by-step explanation:
Given the equations:
1. h(t) = 35 + 16t^2
Comparing the equation above yo the equation of motion,
h(t) = ut + 1/2 × at^2
Where,
h(t) represents the heights in feet, h, of a falling baseball as a function of time, t.
2. h(t) = 6 + 18t - 16t^2
Where,
h represents the heights in feet, h, of the glove of a player leaping up to catch the ball as a function of time, t.
At the time, when the leap time of the glove of the player, t = 0 sec
From equation 1,
h(t) = 35 + 16t^2
= 35 + (16 × 0)
h(0) = 35 ft
This means that the height of the baseball falling is 35 feet at the moment the player begins to catch it.
From equation 2,
h(t) = 6 + 18t - 16t^2
= 6 + (18 × 0) + (16 × 0)
h(0) = 6 ft
This means that the height of the glove of the baseball player as he just leaps to catch the ball is 6 feet.
