Question

A ball is launched straight up with an initial speed of 66mph. The magnitude of the acceleration due to gravity is 9.8 m/s per s(9.8 m/s

2
) which is approximately 22mph per second. Using thismagnitude of 22 mph per second, answer the following questions. When asked for a velocity where sign conveys direction, use the corventional + axis direction as up. 1 2parins: What is the veiocity of the ball is atter launch? −66 miph - A4rimph -22mph 0moh +22mph +44mph +66mph What is the velocity of the ball 2 s after launch? −66mph −44mph −22mph 0mph +22mph +44mph +66mph 2 points What is the velocity of the ball 3 s after launch? −66mph −44mph −22mph 0mph +22mph +44mph +66mph What is the velocity of the ball 4 s after launch? −66mph −44mph −22mph 0mph +22mph +44mph +66mph 2 points What is the velocity of the ball 5 s after launch? −66mph −44mph −22mph 0mph +22mph +44mph +66mph What is the velocity of the ball 6 s after launch? −66mph −44mph −22mph 0mph +22mph +44mph +66mph 2 points How long does it take the ball to reach the highest point? 1 s 2 s 3 s 4 s 5 s 6 s How long does it take the ball to return back down to the same height? 1 s 2 s 3 s 4 s 5 s 6 s

Answer 1

The initial velocity of the ball is 66 mph, which is 29.44 m/s (converting from mph to m/s).

The velocity of the ball after launch is: 29.44 m/s upward or +29.44 m/s.

The velocity of the ball 2 seconds after launch can be calculated using the equation:

v = u + at

where

v = final velocity (unknown)

u = initial velocity (29.44 m/s upward)

a = acceleration due to gravity (-9.8 m/s^2 downward)

t = time (2 s)

Substituting the values, we get:

v = 29.44 - 9.8(2)

v = 9.84 m/s upward or +9.84 m/s

The velocity of the ball 3 seconds after launch can be calculated using the same equation:

v = u + at

where

v = final velocity (unknown)

u = initial velocity (29.44 m/s upward)

a = acceleration due to gravity (-9.8 m/s^2 downward)

t = time (3 s)

Substituting the values, we get:

v = 29.44 - 9.8(3)

v = 0 m/s or 0 m/s upward

The velocity of the ball 4 seconds after launch can be calculated using the same equation:

v = u + at

where

v = final velocity (unknown)

u = initial velocity (29.44 m/s upward)

a = acceleration due to gravity (-9.8 m/s^2 downward)

t = time (4 s)

Substituting the values, we get:

v = 29.44 - 9.8(4)

v = -19.52 m/s or 19.52 m/s downward

The velocity of the ball 5 seconds after launch can be calculated using the same equation:

v = u + at

where

v = final velocity (unknown)

u = initial velocity (29.44 m/s upward)

a = acceleration due to gravity (-9.8 m/s^2 downward)

t = time (5 s)

Substituting the values, we get:

v = 29.44 - 9.8(5)

v = -49.6 m/s or 49.6 m/s downward

The velocity of the ball 6 seconds after launch can be calculated using the same equation:

v = u + at

where

v = final velocity (unknown)

u = initial velocity (29.44 m/s upward)

a = acceleration due to gravity (-9.8 m/s^2 downward)

t = time (6 s)

Substituting the values, we get:

v = 29.44 - 9.8(6)

v = -79.68 m/s or 79.68 m/s downward

To find the time taken by the ball to reach the highest point, we need to use the equation for the time taken for an object to reach its maximum height:

t = u/g

where

t = time taken

u = initial velocity (29.44 m/s upward)

g = acceleration due to gravity (9.8 m/s^2 downward)

Substituting the values, we get:

t = 29.44/9.8

t = 3 seconds

So, it takes the ball 3 seconds to reach the highest point.

To find the time taken by the ball to return back down to the same height, we need to double the time taken to reach the highest point:

t = 2 × 3

t = 6 seconds

So, it takes the ball 6 seconds to return back down to the same height.

Step-by-step explanation: