A ball is thrown straight up at 13.0 m/s near the edge of a 30.0-meter high cliff. The ball lands at the bottom of the cliff. a) What is the acceleration of the ball just after it is thrown (magnitude and direction)? b) What is the acceleration of the ball at the peak of its trajectory (magnitude and direction)? c) What is the velocity of the ball just before impact at the bottom of the cliff? d) How much time after being thrown does the ball take to land?

a/b the acceleration of the ball is g (gravity)

the free-fall equation is
... h = -1/2 g t^2 + 13.0 t + 30.0
... h is the height from the bottom of the cliff
... t is the flight time

d) plug in zero for h to find the flight time

c) the time at peak is on the axis of symmetry of the free-fall equation ... t = -13.0 / -g

subtract the time at peak from the flight time to find the fall time

multiply the fall time by g to find the impact velocity (remember direction and magnitude)

a = g = -9.8 m/s^2.

b. a = g = -9.8 m/s^2.

c. V = Vo + g*Tr = 0.
13 - 9.8Tr = 0
Tr = 1.33 s. = Rise time.

h = ho + Vo*Tr + 0.5g*Tr^2.
h = 30 + 13*1.33 - 4.9*1.33^2 = 38.6 m.

V^2 = Vo^2 + 2g*h.
V^2 = 13^2 - 19.6*38.6 = 925.6
V = 30.4 m/s.

d. h = 0.5g*Tf^2 = 38.6.
4.9*Tf^2 = 38.6
Tf = 2.81 s. = Fall time.

Tr + Tf = 1.33 + 2.81 = 4.12 s. = Time in flight.

To answer these questions, we need to apply the laws of motion and equations of motion. Let's take it step-by-step:

a) What is the acceleration of the ball just after it is thrown (magnitude and direction)?
The initial velocity of the ball is 13.0 m/s upwards. As the ball is thrown straight up, it experiences a constant acceleration due to gravity. The acceleration due to gravity on Earth is approximately 9.8 m/s² downward. So, the acceleration of the ball just after being thrown is 9.8 m/s² downward.

b) What is the acceleration of the ball at the peak of its trajectory (magnitude and direction)?
At the peak of its trajectory, the ball momentarily comes to a stop before falling back down. This means its velocity is zero and its acceleration is solely due to gravity. Therefore, the acceleration at the peak of its trajectory is still 9.8 m/s² downward.

c) What is the velocity of the ball just before impact at the bottom of the cliff?
To find the velocity just before impact, we can use the equation of motion for vertical motion: v² = u² + 2as, where v is final velocity, u is initial velocity, a is acceleration, and s is displacement.
In this case, the initial velocity is 13.0 m/s upwards, the acceleration is 9.8 m/s² downward, and the displacement is the height of the cliff, which is 30.0 meters.
Plugging the values into the equation, we get v² = (13.0 m/s)² + 2*(-9.8 m/s²)*(-30.0 m).
Simplifying the equation, v² = 169.0 m²/s² + 588.0 m²/s².
Adding the values, we get v² = 757.0 m²/s².
Taking the square root of both sides, we find that the velocity just before impact is approximately 27.5 m/s downward.

d) How much time after being thrown does the ball take to land?
To calculate the time taken by the ball to land, we can use the equation of motion: v = u + at, where v is final velocity, u is initial velocity, a is acceleration, and t is time.
In this scenario, the final velocity is 0 m/s (when it lands), the initial velocity is 13.0 m/s upwards, and the acceleration is 9.8 m/s² downward.
Plugging the values into the equation, we get 0 = 13.0 m/s + (-9.8 m/s²)t.
Simplifying the equation, -13.0 m/s = -9.8 m/s²t.
Dividing both sides by -9.8 m/s², we get t = 1.33 seconds.
So, the ball takes approximately 1.33 seconds after being thrown to land.