Question

Q3. A ball is thrown vertically upward and returns to the thrower in 20 seconds Calculate the
velocity with which it was thrown and the maximum height attained by the ball (use G = 10
mls)

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Answer 1

QUESTION:

A ball is thrown vertically upward and returns to the thrower in 20 seconds Calculate the velocity with which it was thrown and the maximum height attained by the ball (use g = 10m/s)

ANSWER:

maximum height = 500 m

maximum height = 500 mvelocity with which it was thrown  = 100 m/s

Explanation:

given that,

A ball is thrown vertically upward

here,

let the velocity at which ball is thrown be u

and

also given that,

the ball returns to the thrower in 20 seconds,

we know that,

time taken to the ball to go the maximum height = the time taken to fall it down

so,

time taken by ball to go maximum height = 20/2

= 10 seconds,

since,

ball will stop at the maximum height for a moment

so ,

final velocity = 0

now,

its given that,

initial velocity = u

final velocity(v) = 0

time(t) = 10 seconds

gravitational acceleration = 10 m/s²

now,

by the equation of gravitation

v = u + gt

putting the values,

0 = u + 10(10)

u + 100 = 0

u = -100 m/s

negation of velocity shows the velocity against the gravitation

so,

___________________

velocity with which it was thrown

= 100 m/s

___________________

now,

its given that,

initial velocity(u) = 100 m/s

final velocity(v) = 0

time(t) = 10 seconds

gravitational acceleration(g) = 10 m/s²

let the maximum height at which ball will reach be h

by the equation of gravitation

v² = u² + 2gh

putting the values,

(0)² = (100)² = 2(10)h

20h + 10000 = 0

20h = -10000

h = -10000/20

h = -500 m

negation of height shows the height against the gravitation

so,

__________________

maximum height = 500 m

__________________

velocity with which it was thrown  = 100 m/s

__________________

Answer 2

We know that time of descending = time if ascending

total time taken=20 then time for ascending=10sec

at highest point v=0

g=-10m/s^2(for ascending)

By using first eq of motion

v=u+gt

-u=-gt

u=10×10

u=100m/s

Max height attained

by using third eq of motion

v^2=u^2+2as

u^2=2as

100×100=2×10×s

s=500m