Question

A ball is thrown vertically upwards with an initial velocity of 49 m/s.
Calculate :
(i) the maximum height attained,
(ii) The time taken by it before it reaches the ground again. ( Take g = 9.8m/s²).​

Answer 1

Answer:

i. 122.5m

ii. 10s

Explanation:

Given:- u=49 m/s. ,. v=0 ,. a/g=9.8 m/s².

i. by third formula of motion(in case of gravitation)

v²=u²-2gh

0= (49)² - 2×9.8h

-2401= -19.6h

h = -2401/-19.6

h = 122.5m

ii. by second formula of motion (in case of gravitation)

t = 2u/g

t = 2×49/9.8

t = 98/9.8

t = 10s

Answer 2

Explanation:

intial velocity of ball=49m/s

gravitational force=9.8

maximum height attained=u^2/2g(by 3rd equation of motion v^2-u^2=2gh derive this formula

=(49)^2/2*9.8

• =2401/19.6
• =122.5m
• maximum height attained=122.5
• time taken to reach the ground again
• height. =122.5
• g=9.8m/m^2
• as ball is dropped freely u=0
• from equation of motion h=gt^2
• 122.5=1/2*9.8*t^2
• 122.5*2=9.8*t^2
• 245=9.8*t^2
• 245/9.8=t^2
• 25=t^2
• √25=t
• 5=t
• time taken to reach the ground=5mls