Question

a ball thrown up vertically returns to the thrower after 6 seconds find ,a.the velocity with which it was thrown up ,b.the maximum height it reach and,c. its position of 4 second​

Answer 1

Answer:

Time = 2u/g

6 = 2u/10

u = 30m/s

b) H = u2/2g

H = 900/20

H = 45m

c) Sn = u + g/2(2n -1)

Answer 2

Questioñ:

A ball thrown up vertically returns to the thrower after 6 seconds. Find

a) The velocity with which it was thrown up,

b) The maximum height it reaches and

c) Position after 4 seconds.

Solutioñ:

• Time of ascent is equal to time of descent the vertex a total of 6 second for its upward and downward journey.

Hence,

• It has taken 3s to attain the maximum height.

• Final velocity of the ball at the maximum height,v=0

• Acceleration due to gravity, g=-9.8ms^-2

From the first equation of motion,

v=u+vt

0=u+(-9.8×3)

u=9.8×3=29.4 ms^-1

Hence,

The ball was thrown upward with a velocity of 29.4 ms^-1.

• let the maximum height attained by the ball be h

Initial velocity during the upward journey,u=29.4 ms^-1.

Final velocity,v=0

Acceleration due to gravity ,g=-9.8 ms^-2.

From the second equation of motion,

s=ut+1/2×gt²

h=29.4×3+1/2×(-9.8)×3²=44.1m

• Ball attains the maximum height after 3 seconds, after attaining this height, it will start falling downwards.

In this case,

Initial velocity,u=0

Distance travelled by its it during its downward journey is remaining 1 second is given by

s=0×t+1/2×9.8×1²=4.9

Total height=44.1 m

This means that the ball is 39.2 (44.1 - 4.9m) above the ground after 4 seconds.

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