Map sapling learning Bob has just finished climbing a sheer cliff above a beach,
ID: 1394578 • Letter: M
Question
Map sapling learning 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 32.5 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 128 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? Number 32.5 m/s A. 0.710 s 2 m 128 mExplanation / Answer
in half of time= 0.710 sec the ball will be at maximum height. at maximum height Vy = 0. so
Vy = uy + ay*t ( Vy = final velocity in vertical direction, uy = initial velocity in vertical direction)
0 = u*sin@ - gt ( @= angle with which balll was thrown, a= -g)
u*sin@ = gt ( t= 0.710/2 sec)
@ = sin-1 ( 9.8 * 0.710/2) / 32.5]
@ = 6.1450
initial horizontal velocity ux = u*cos@
= 32.5*cos ( 6.145) = 32.3 m/s
so total time of flight will be T. by equation of motion in horizontal direction:
Sx = ux* t + (1/2) * ax*t2
128 = 32.31* T +0 ( ax = acceleration in horizontal direction = 0)
T = 3.96 sec
vertical displacement from launch height to base of cliff :
sy = uy * t + (1/2) * ay*t2
h = u*sin@*t + ( 1/2) * (-g)*t2
h = 32.5*sin ( 6.145)*3.96 - (1/2)*9.8*3.962
h = - 63.06 m
-ve sighindicates that this is fall in height.
so cliff height ( how heigh is bob), H = h-2 = 63.06-2
H = 61.06 m