sharon is jumping from an 18-foot diving board with an initial upward velocity of 4 ft/s. when sharon jumps…
sharon is jumping from an 18-foot diving board with an initial upward velocity of 4 ft/s. when sharon jumps, megan throws a beach ball up to sharon with an initial upward velocity of 16 ft/s from a height 5 feet off the ground. to the nearest hundredth of a second, how long after she jumps does the ball reach sharon?\n0.65 seconds\n0.92 seconds\n1.08 seconds\n1.15 seconds
Answer
Explanation:
Step1: Recall the height formula
The height of an object in vertical motion is given by ( h(t) = -16t^2 + v_0t + h_0 ), where ( v_0 ) is the initial velocity and ( h_0 ) is the initial height.
For Sharon: ( h_S(t) = -16t^2 + 4t + 18 ) (initial height 18 ft, initial velocity 4 ft/s)
For the beach ball: ( h_B(t) = -16t^2 + 16t + 5 ) (initial height 5 ft, initial velocity 16 ft/s)
Step2: Set heights equal
To find when the ball reaches Sharon, set ( h_S(t) = h_B(t) ):
( -16t^2 + 4t + 18 = -16t^2 + 16t + 5 )
Step3: Simplify the equation
Add ( 16t^2 ) to both sides:
( 4t + 18 = 16t + 5 )
Subtract ( 4t ) from both sides:
( 18 = 12t + 5 )
Subtract 5 from both sides:
( 13 = 12t )
Step4: Solve for t
Divide both sides by 12:
( t = \frac{13}{12} \approx 1.0833 )
Rounding to the nearest hundredth, ( t \approx 1.08 ) seconds.
Answer:
1.08 seconds (corresponding to the option "1.08 seconds")