Question

A baseball is hit from a height of 3.5 feet above the ground with an initial upward velocity of 48 feet per second and an initial outward velocity of 85 feet per second. Write the function h(t) that gives the height of the ball after time (t).

Answer 1

this is similar to the first question with differentt values

h(t) = 48t - 16t^2 + 3.5

Answer 2

A baseball is hit from a height of 3.5 feet above the ground with an initial upward velocity of 48 feet per second and an initial outward velocity of 85 feet per second. How long will it be before the ball hits the ground? Round your answer to the nearest thousandth of a second.

Answer 3

48t + 9.8t^2=3.5
solve this

Answer 4

Would I use the quadratic formula?

Answer 5

naren the question is using feet not metres so g is not 9.8 but 32 ft/s/s

Answer 6

sorry i had to go again hold on a sec

Answer 7

O.k no problem!

Answer 8

Just for clarification the question is: How long will it be before the ball hits the ground?

Answer 9

yes use 48t - 16t^2 +3.5 = 0

u can use the formula to solve this - i use my calculator because i'm lazy!!

the solution is 3.071 seconds (the other solution is -0.07 which u ignore)

Answer 10

O.k that's what I thought. Thanks! Now...
A baseball is hit from a height of 3.5 feet above the ground with an initial upward velocity of 48 feet per second and an initial outward velocity of 85 feet per second. How far will the ball travel before it hits the ground?

Answer 11

right we use the time taken = 3.071 seconds and multiply by the horizontal speed which is 85
distance = 3.071 * 85 = 261.04 ft

Answer 12

Oh o.k I see. Thank you for explaining. Now...
A baseball is hit from a height of 3.5 feet above the ground with an initial upward velocity of 48 feet per second and an initial outward velocity of 85 feet per second. What is the maximum height the ball will reach?

Answer 13

the equation of motion for the vertical component is

v^2 = u^2 + 2gh at maximum height the speed v = 0 u = initial speed = 48
and g =-32 ft/s/s , h = height

0 = 48^2 +2 * -32 * h
h = 2304 / 64 = 36 ft

add initial height of 3.5

= 39.5 ft

Answer 14

O.k I understand. LAST QUESTION:
A baseball is hit from a height of 3.5 feet above the ground with an initial upward velocity of 48 feet per second and an initial outward velocity of 85 feet per second. After how long will the ball be in the air when it is on its way down at a height of 15 feet? Round to the nearest thousandth of a second.

Answer 15

48t - 16t^2 + 3.5 = 15

solving for t

t = 2.737 secs

Answer 16

must go - take care

Answer 17

THANK YOU MUCH! You too!