In: Physics
Bob has just finished climbing a sheer cliff above a beach, and wants to figure out how high he climbed. All he has to use, however, 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 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 0.910 seconds the ball is once again level with Bob. Bob can\'t see well enough to time when the ball hits the ground. Bob\'s friend then measures that the ball landed 125 m from the base of the cliff. How high up is Bob, if the ball started from exactly 2 m above the edge of the cliff?
Given that
The initial speed ofthe ball (u) =33.7m/s
The time of flight for the ball(t) =0.910s
Acceleration due to gravity (g) =9.81m/s2
The time for which the ball remained in the air along the tajectory is given by
t =2usintheta/g
Then angle with which it is projected is given by
sintheta =gt/2u
theta=sin-1(gt/2u) =sin-1(9.81*0.910/2*33.7) =7.61degrees
Now the horizontal component of velocity is given by
ux =ucosthea =(33.7)cos(7.61) =33.403m/s
and the vertical component is given by
uy =usintheta =(33.7)sin(7.61)=4.462m/s
Then the time taken for the ball to reach 125m from the projection is gvien by
x =ux*t
t =x/ux =125/33.403 =3.742s
How high up is Bob, if the ball started from exactly 2 m above the edge of the cliff is given by
y =uyt+(1/2)at2
y =-(4.462)(3.742)-0.5(9.81)(3.742)2=-85.372m
How high up is Bob, if the ball started from exactly 2 m above the edge of the cliff is =85.372-2m =83.372m