Question

1) Bob has just finished climbing a sheer cliff above a level beach and wants to figure out how high he climbed. All he has to use is a baseball, a stopwatch, and a friend on the ground below with a long measuring tape. Bob is a pitcher and knows that the fastest he can throw the ball is about ?0=33.7 m/s.Bob starts the stopwatch as he throws the ball, with no way to measure the ball's initial trajectory, and watches carefully. The ball rises and then falls, and after ?1=0.910 s. the ball is once again level with Bob. Bob cannot see well enough to time when the ball hits the ground. Bob's friend then measures that the ball hit the ground ?=125 m. from the base of the cliff.

How high above the beach was the ball when it was thrown?

2) A planet of mass ?=4.05×10^24 kg is orbiting in a circular path a star of mass ?=1.75×10^29 kg. The radius of the orbit is ?=4.35×10^7 km.

What is the orbital period (in Earth days) of the planet ?planet?

Tplanet = days

Answer 1

Question 1

Answer: 51.86 m

Solution-

In the vertical direction, we can say the ball went to a height in half the time given to go up and down, and it will momentarily stop at the height under the acceleration of gravity

Therefore apply vf = vo + at

0 = vo + (9.8)(.910)/2

vo = 4.459 m/s

That is the y velocity initial

We can then apply Vy = sin(angle)(v)

4.459 = 33.7(sin angle)

angle = 7.59 degrees

The x component of the velocity will be 33.7(cos 7.59)

Vx = 33.4 m/s

Then d = vt to find the time to travel the horizontal distance

125 = 33.4(t)

t = 3.74 sec

That time is used to travel downward to the ground

d = vot + 0.5*a*t^2

d = (4.459)(3.74) + (.5)(-9.8)(3.74)2

d = -51.86 m

Question 2

Answer: 192.95 days

Solution-

               centripetal force = gravitational force

mw^2R=mMG/R^2

w=2π/T

then

m(2π/T)^2R=mMG/R^2

from which

T=2π[R^3/(MG)]^1/2

Now Putting the Values

T=2*3.14[(4.35x10^10)^3 / (1.75x10^29*6.67x10^-11)]^1/2

T=1.6676x10^7 s = 192.95 days