Question

A diver launches himself upwards from a springboard at a speed of 6.5 m s−1. His trajectory is (practically) vertical and he strikes the water 1.8 s after leaving the board. How high above the water is the springboard?

Answer 1

we have given parameters as follows

inital speed u = 6.5 m s−1

total time to reach the water T =  1.8 s

now here as stated diver launches himself upward first so now first calculate the time he spend above springboard

form the first law of motion we have

v = u + gt ....................... 1

here g = 9.81 ms-2 and will be negative as body is moving upward

v = 0 as at highest point kinetic energy of the body will be zero\

so that

0 = 6.5 - 9.81*t1

so from here we will get t1 which will be as

t1 = 0.66258 s

now during this time the distance travelled by the diver in upward direciton will be as

S1 = ut1 - 0.5*g*ti^2

here S1 is the upward distance travelled by the diver

now on plugging the respective values we will get

S1 = 2.1534 m

now at the highest point the velocity of the diver is zero which we can treat as initial velovity for the next case and the time we have remaining as

t2 = T-t1 = 1.13742 s

so form here

lets find the velocity with which the diver strikes the water which is given as

v2 = u2 + gt2

here u2 = 0 as explianed above so that

v2 = 11.15809 m/s

now the distance travelled from the top position will be given as

v2^2 = u2^2 + 2gS2

from here we can get the S2

S2 = (11.15809)^2/2*9.81 = 6.34571725 m

hence the height of the springboard will be

S = S2-S1 = 6.34571725 - 2.15347 = 4.20 m

hence  springboard is 4.20 m high above the water.