A diver springs upward with an initial speed of 1.6 m/s from a 4.5-m board. ?

(a) Find the velocity with which he strikes the water. [Hint: When the diver reaches the water, his displacement is y = -4.5 m (measured from the board), assuming that the downward direction is chosen as the negative direction.]
_m/s

(b) What is the highest point he reaches above the water?
_m

3 Answers
Jane
Mar 5, 2018

A). Simply the diver after jumping,first go upwards then come back to the point of the jump,at this point he regains the velocity with which he jumped upwards i.e #1.6 ms^-1# but the direction will be downwards.

So,we can say,if he strikes the water with a velocity of #v#,then we can write,

#v^2 =1.6^2 + 2g 4.5# (using, #v^2=u^2+2gh#)

so, #v=9.527 ms^-1#

B) Now,suppose he went o a maximum height of #h# upwards from his point of jump,so at the highest point his velocity wil become zero,so we can apply the same equation to calculate #h#,

So, #0^2 =1.6^2 -2gh#

or, #h=0.131m#

So,above the water he reaches a maximum height of #4.5+0.131=4.631m#

Jane
Mar 5, 2018

#A) 9.527 ms^-1#

#B) 4.631 m#

Explanation:

Alternatively,you could have applied law of conservation of energy to solve this.

At his jumping point,he has total energy of #mgh + 1/2 mv^2# (where, #m# is his mass, #h# is the height of the jumping point above the water and #v# is his velocity)

Now,if ihe reaches a maximum height of #H# above the water,at that point he will have only potential energy i.e #mgH#

So,from conservation of energy we can equate both,

#mgh +1/2 mv^2 =mgH#

putting the value of #g# and #v=1.6ms^-1# we get, #H=4.631m#

Now,when he strikes water,he has purely kinetic energy,and if at that poin he has a velocity of #V# then we can write,

#mgh + 1/2 mv^2 = 1/2 mV^2#

solving we get, #V=9.527ms^-1#

A08
Mar 5, 2018

Alternate answer, as per hint given in the question.

Explanation:

(a). Let the origin be located at the place from where diver jumps. Noting that gravity #g=9.81\ ms^-2# acts in the downwards direction, opposite to the initial velocity, applicable kinematic expression is

#v^2-u^2=2gh# .......(1)

Inserting given values we get

#v^2 -(1.6)^2 = 2(-9.81)(-4.5)#
#=>v=9.5\ ms^-1#

(b) At the highest point his velocity will be zero. Inserting given values in (1) we get

#0^2 -(1.6)^2 =2(-9.81)h_0#
where height #h_0# is from the board
#=>h_0=0.13\ m#

He reaches highest of

#4.5+0.13=4.63\ m# above the water

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