Question

A ball is projected vertically upward with a speed of 50 m/s. Find
(a)The maximum height
(b) Time taken to reach the maximum height
(c) The speed at half the maximum height, take g=10 m/s.​jaldi bata do

Answer 1

Answer:

Explanation:

Solution :

Initial speed= u=50m/s

g=10m/s2

At maximum height V=0 m/s

Maximum height:

V²-u²=2as

0²-50²=2*-10*s

s=50*50/20

s=125m

∴maximum height reahed =125m

b) time taken to reach maximum height

t=u/g

t=50/10=5 sec

∴Time taken to reach maximum height =5 sec

c) the speed at half the maximum height:

s=125/2=62.5m

v=?

u=50m/s

a=-10m/s²

From  v²-u²=2as

V=√u²+2as

V=√50²+2x(-10)x 62.5

v=√2500-1250

v=√1250=35 m/s

Answer 2

PHYSICS

A ball is projected vertically upward with a speed of 50 m/s. Find (a) the maximum height, (b) the time to reach the maximum height, (c) the speed at half the maximum height. Take g = 10m/s

2

.

December 20, 2019avatar

Sristi Kanojia

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ANSWER

(a) Initial velocity of ball (u)=50m/s, acceleration of ball =−g,

final velocity at the highest point (v)=0

So applying the 3rd equation of motion we get:

v

2

=u

2

−2gh

max

⇒0=50

2

−2×10×h

max

⇒h

max

=

20

2500

=125m

(b) Let the time required to reach max height be t. Then applying 1st equation of motion we get:

v=u−gt

⇒0=50−10t

⇒t=5s

(c) Let speed at half of max height be V then:

V

2

=50

2

−2g

2

125

⇒V

2

=2500−1250=1250

⇒V=

1250

=35.35m/s.