A ball is thrown from the edge of a cliff with an initial velocity of 60·m/s upward. Answer the following questions using + (upward) and - (downward) to indicate the direction of the velocity. Ignore air resistance and assume the ball does not hit the ground during the question.

(a) Find the velocity of the ball 5 seconds after it is thrown. 1 m/s.

(b) Find the velocity of the ball 10 seconds after it is thrown. 2 m/s.

(c) How high does it rise above the edge of the cliff (from where it was thrown)? 3 m.

To solve this problem, we can use the equations of motion for linear motion with constant acceleration. In this case, the acceleration is due to gravity and is approximately 9.8 m/s^2 downward.

(a) To find the velocity of the ball 5 seconds after it is thrown, we can use the equation:

v = u + at

where:
v = final velocity
u = initial velocity
a = acceleration
t = time

Given:
u = 60 m/s (upward)
a = -9.8 m/s^2 (downward)
t = 5 s

Substituting the values into the equation, we have:

v = 60 + (-9.8)(5)
v = 60 - 49
v = 11 m/s

Therefore, the velocity of the ball 5 seconds after it is thrown is +11 m/s (upward).

(b) To find the velocity of the ball 10 seconds after it is thrown, we can again use the equation:

v = u + at

Given:
u = 60 m/s (upward)
a = -9.8 m/s^2 (downward)
t = 10 s

Substituting the values into the equation, we have:

v = 60 + (-9.8)(10)
v = 60 - 98
v = -38 m/s

Therefore, the velocity of the ball 10 seconds after it is thrown is -38 m/s (downward).

(c) To find the height the ball rises above the edge of the cliff, we can use the equation for displacement:

s = ut + (1/2)at^2

Given:
u = 60 m/s (upward)
a = -9.8 m/s^2 (downward)
t = time taken to reach maximum height

At the maximum height, the velocity of the ball becomes zero. Therefore, we can set v = 0 and solve for t:

v = u + at
0 = 60 + (-9.8)t
9.8t = 60
t = 60 / 9.8
t ≈ 6.12 s

Now, we can substitute the values into the equation for displacement:

s = (60)(6.12) + (1/2)(-9.8)(6.12)^2
s ≈ 366.72 - 180.205
s ≈ 186.515 m

Therefore, the height the ball rises above the edge of the cliff is approximately 186.515 m.

To answer these questions, we can use the laws of motion, specifically the equations of motion for a body in free fall under constant acceleration.

The equation we'll be using is:
v = u + at

where:
v = final velocity
u = initial velocity
a = acceleration
t = time

For (a), we need to find the velocity of the ball 5 seconds after it is thrown, given that the initial velocity is 60 m/s upward. Since the ball is thrown upward, the initial velocity is positive. However, due to gravity, the acceleration acting on the ball is negative. Let's substitute the values into the equation:

v = u + at
v = 60 m/s + (-9.8 m/s^2) * 5 s
v = 60 m/s - 49 m/s
v = -11 m/s

So, the velocity of the ball 5 seconds after it is thrown is -11 m/s. The negative sign indicates that the velocity is downward.

For (b), we need to find the velocity of the ball 10 seconds after it is thrown. Using the same equation, let's substitute the given values:

v = u + at
v = 60 m/s + (-9.8 m/s^2) * 10 s
v = 60 m/s - 98 m/s
v = -38 m/s

Therefore, the velocity of the ball 10 seconds after it is thrown is -38 m/s. Again, the negative sign indicates that the velocity is downward.

For (c), we need to determine how high the ball rises above the edge of the cliff. To do this, we can use another equation of motion:

s = ut + (1/2)at^2

where:
s = displacement or height
u = initial velocity
a = acceleration
t = time

Since the ball was thrown upward, the initial velocity is positive, and the acceleration due to gravity is negative. Let's plug in the values:

s = ut + (1/2)at^2
s = 60 m/s * 5 s + (1/2) * (-9.8 m/s^2) * (5 s)^2
s = 300 m - (1/2) * 9.8 m/s^2 * 25 s^2
s = 300 m - 122.5 m
s = 177.5 m

Hence, the ball rises 177.5 meters above the edge of the cliff.

call g = -10m/s^2

v = vi - g t = +60 - 10t
when t = 5
v = 60 -10(5) = 10 m/s

when t = 10
v = 60 - 10 (10) = -40 m/s

y = yi + vi t - (1/2)g t^2
what is t at the top where v = 0 ??
0 = 60 - 10 t
t = 6 seconds to top
y = 0 + 60(6) -5 (36)
= 180