Question

1 An object is thrown up with a velocity of 98 m/s. Find the maximum

   height it reaches and also find the total time taken to return to the  

   starting point.  

explanation need that is short with step​​

Answer 1

Answer:

ANSWER

Here, u=98m/s, t

1

=7s. When body reaches maximum height, its velocity becomes v

1

=0.

Maximum height reached can be obtained using:

v

2

=u

2

+2ah

h

max

=

2g

u

2

Now considering initial velocity as v

1

=0, displacement as s=

2g

u

2

,

We can find time for the body to come back from the maximum height using:

s=ut+

2

1

at

2

2g

u

2

=

2

1

gt

2

t=

g

u

Thus, total time, t=

g

2u

=20 s

Now the time it takes to reach point P from ground and to reach from point P to ground is same i.e. 7s each time.

Thus, it takes 20s−7s=13s to come back to the same point P since its projection.

Answer 2

Answer:

height = 490m and time = 20 sec

Step-by-step explanation:

u = 98m/s

v = 0m/s

a = g = 9.8m/s^2

s = ?

time to return = ?

for s we will apply 3rd equation of motion,

i.e. v^2 - u^2 = 2as

so, 0 - 98 × 98/ 2 ×(-9.8) = s [-ve because going up]

s = 98 × 98/ 2 ×9.8

s = 98×10 / 2

s = 490m

Now time of ascend is equal to time of descend

so, total time = time of going up × 2

For calculating time of ascend, apply 1st equation of motion,.i.e. v = u + at

so,

0 = 98 + (-9.8)t

-98/ -9.8 = t

t = 10sec

Now total time(T)

T = t × 2

T = 10 ×2

T = 20 sec