Question

Using Integral Calculus
A ball is thrown downward from a window that is 80 m above the ground with an initial velocity of -67meter/second;
(a) when does the ball strike on the ground, and
(b) with what speed will it strike on the ground.

Answer 1

please help me

Answer 2

g is approximately 9.81 m/s^2

Answer 3

what it means g^w in above?

Answer 4

g stands for the acceleration due to gravity , which is a constant on the surface of earth

Answer 5

On the surface of earth acceleration due to gravity is about 9.81 m/s^2
Choosing up is positive

d^2y/dt^2 = -9.81 m/s^2

integrating we have

dy/dt = -9.81t + C

y ' (0) = -67

-67 = -9.81*0 + C

-67 = C

therefore

dy/dt = -9.81*t - 67

integrating again we have

y(t) = -9.81t^2/2 - 67t + C

y(0)=87

87 = -9.81*(0)^2/2 - 67(0) + C

87 = C

y(t) = -9.81*t^2/2 -67t + 87

Answer 6

i can't get it what is the second velocity and the time does the ball strike on the ground

Answer 7

i thought if we estimate the gravity it must be 9.82 m/s^2

Answer 8

the second derivative of position (also known as acceleration) is constant -9.81

Answer 9

you can use 9.82 if you want

Answer 10

ahhh okay, thank you so much

Answer 11

why is it 87 is C?

Answer 12

i can't get it why c is equal to 87.

Answer 13

oh that was a typo

Answer 14

On the surface of earth acceleration due to gravity is about 9.81 m/s^2
Choosing up is positive

d^2y/dt^2 = -9.81 m/s^2

integrating we have

dy/dt = -9.81t + C

y ' (0) = -67

-67 = -9.81*0 + C

-67 = C

therefore

dy/dt = -9.81*t - 67

integrating again we have

y(t) = -9.81t^2/2 - 67t + C

y(0)=80

80 = -9.81*(0)^2/2 - 67(0) + C

80 = C

y(t) = -9.81*t^2/2 -67t + 80

Answer 15

thanks.