Question

If a ball is thrown vertically upwards with speed 40 m/s, the distance covered during the last second of its descent will be (g = 10 m/s)

70 m

90 m

35 m

125 m​

Answer 1

Answer:

90m. I think..........

Answer 2

Given : a ball is thrown vertically upwards with speed 40 m/s,  

g = 10 m/s

To Find : distance covered during the last second of its descent

70 m

90 m

35 m

125 m​

Solution:

ball is thrown vertically upwards from Ground

Speed = 40 m/s

Time taken to Reach peak point  = T

V = U + at

a = - g = - 10

=> 0 = 40 - 10T

=> T = 4 sec

Hence total travel time  =  8 sec

Distance covered in last second of its descent

= Distance covered between 7th and 8th sec.

Velocity after 7 sec = 40 - 10(7) = - 30 m/s

Velocity after 8 sec = 40 - 10(8) = - 40 m/s

Distance covered using V² - U² = 2as

=> (-40)² - (-30)² = 2(10) S

=> 700 = 20S

=> S = 35 m

Another Way

Distance covered in last 1 sec = Distance covered in 1st sec

S = Ut + (1/2)at²

= 40*1 +(1/2)(-10)1²

= 40 - 5

= 35 m

the distance covered during the last second of its descent = 35 m

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