Please help me solve the question Bob has just finished climbing a sheer cliff a
ID: 584245 • Letter: P
Question
Please help me solve the question
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 0.710 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 124 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?Explanation / Answer
from the figure
for vertical motion
uy=u*sin(theta)
t=TH
ay=-g
vy=0
from the first equation of motion
vy=uy+ay*t
0=u sin(theta)+(-g)*TH
TH=(u*sin(theta))/g
2*0.710 = (33.7*sin(theta))/9.8
sin(theta)=(1.42*9.8)/33.7
sin(theta)=0.41
theta =24.20 deg
from the third equation of motion
(vy)^2=(uy)^2+2*ay*sy
vy=0
uy=u*sin(theta)
sy=Hmax
ay=-g
0=((u*sin(theta))^2)+2*(-g)*Hmax
Hmax=((u*sin(theta))^2)/2g
Hmax=((33.7*0.41)^2)/2*9.8
Hmax=9.72 m
from the second equation of motion
ux=u*cos(theta)
ax=0
Sx=R
Sx=ux*t+1/2*ax*t^2
R=u*cos(24.20)*t+0
124=33.7*cos(24.20)*t
t=6.17 s
for the vertical motion
from the second equation of motion
sy=(uy*t)+1/2*ay*(t^2)
given
uy=0
t=T
ay=-g
sy=-h
puting values in equation
T=sqrt((2*h)/g)
6.17=sqrt((2*h)/9.8)
38.12=(2*h)/9.8
h=186.79 m
now total height
Htotal=Hmax+h+2
Htotal=9.72+186.79+2
Htotal=198.51m