Frank A. answered • 02/17/20
Knowledge, experience and insight from decades of teaching physics.
Using the standard symbols,
h=vi•t + 0.5•g•t2
Since the object is falling, vi = 0.
rearranging, t=√(2•h/g)
Gabrielius T, if you need further clarification about this or any other problem, please feel free to ask Mr. A
Marvellous A.
Hi sir....in a situation where vi=40m/s and h=50m and we are asked to find the range....how do we do that??.....will t=√2H/g ????please I need an urgent response07/19/20
Frank A.
No that is only true when Vi=0. For the question you are now asking, i will need the complete question to be able to help. Mr. A07/19/20
Tsoy G.
Hi Mr. A, Can you explain to me how did you come up with the 2 within the sqrt(2h/g)09/21/20
Frank A.
Use kinematics equation x=vit +1/2at^2, replace x=h, a=g, vi=0. 2h=gt^2. t^2=2h/g09/21/20
MARTA C.
thx!02/14/22
Shahin K.
Marvellous A. The question is of horizontal projectile motion...so we take time as (√2h/g) because it's dropping downwards and doesn't depend upon Vi.. But range is the distance covered in x axis..so Range depends upon Vi on x axis.. T=√2h/g=√2×50/10=√10 ; Vi=40 ; Range(basically dist covered on x axis)=vi×t =40√1007/08/22
Anamika R.
Can I use this equation to find t when a balls is thrown downwards with velocity u from a tower of height h08/29/22
Sher A.
Hi sir Q.. time required by the projectile to reach the summit point.. Famula t=√2h/g ? How sir? If we use time and height formula for projectile motion please sir09/02/22
Gabrielius T.
Thank you.02/18/20