Question

A baseball is thrown from a height of 17 feet with an initial velocity of 55 ft/sec at an angle of 75 degrees with horizontal.

Answer 1

what is the question?

Answer 2

write parametric equations of motion

Answer 3

estimate how long ball was in air

Answer 4

estimate max. height of ball
estimate how far ball was thrown

Answer 5

Thrown at 55 ft/s at an angle of 75°

Vertical component = 55 sin 75° = 53.1259 ft/s
Horizontal component = 55 cos 75° = 14.2350 ft/s

let t = time in seconds
let d = horizontal distance in feet
let h = vertical height in feet

a.)

d = 14.2350t
h = -16t^2 + 53.1259t + 17


b.)

Set h = 0 and solve for t

0 = -16t^2 + 53.1259t + 17
t = -0.2940, 3.6143

Since the ball can't hit the ground before it is thrown, reject the negative answer. t = 3.6143 seconds


c.)

h = -16t^2 + 53.1259t + 17 represents a parabola opening down. The vertex will be at
x = -b/2a. In this case, -53.1259/-32 = 1.6602

Sub the above time into the equation for h. At 1.6602 seconds, the ball will be at it's maximum height of 61.0994 feet.

Since the ball was in the air for 3.6143 seconds, according to the equation for d, it traveled
14.2350 * 3.6143 = 51.4496 feet.

Answer 6

wow thank you so much

Answer 7

no problem

Answer 8

however i have one more if ur willing to help again sir

Answer 9

i can try

Answer 10

the center-field fence in ballpark is 20 feet high and 550 feet from home plate. the baseball is hit 2.5 feet above ground. It leaves the bat at an angle of THeta degrees with horizontal speed of 100 feet per second.

Answer 11

write parametric equations that model flight of baseball if theta is 15 degrees. is home run hit

Answer 12

also if theta is 23 degrees