An object can be thrown vertically upwards or at an angle to the ground, in both cases it is needed that time of flight be 5 seconds. This means it's time of ascent (going up) is 2.5 seconds and time of descent(coming down) is also 2.5 seconds. So, it reaches highest point 2.5 seconds after it is thrown. At highest point the vertical component of velocity1 of the object becomes zero for an instance.
Now, kinematics equation can be used to solve this.
v = uy - g*t
where v is final velocity at top =0. uy is initial vertical velocity1. g is accleration due to gravity(9.8ms-2). t is time of ascent.
putting values we get.
0 = uy - 9.8 * 2.5
or uy = 24.5 ms-1
So we need to throw an object with vertical velocity = 24.5 meters per second so that it remains in air for 5 seconds.
1. If object is thrown at an angle then vertical component of velocity of projection is taken.
If object is thrown vertically upwards then vertical component of velocity of projection is
same as velocity.
vertical component of velocity of projection(uy) = u*sin(θ), where u is velocity of projection
and θ is angle of projection with respect to horizontal.
2. Accleration due to gravity is different for different places. 9.8ms-2 is an approximate
value.
3. Here, air resistance and wind speeds have been neglected as they make the calculation very
tedious and they are always varying from time to time and place to place.
I have to make two assumptions:-- The object was not thrown down. Its speed was zerojust as it left the dropper's hand.-- There is no air in this question.With those assumptions . . .H = 1/2 G T2T = sqrt( 2H/G )T = sqrt( 2 x 176.4/9.8 ) = 6 seconds
yes, i just test right now
False, provided the drop occurs no sooner than the throw, and the ground is flat .
No. They both hit the ground at the same time, because the VERTICAL component of velocity in both cases is the same.
No. They both hit the ground at the same time. This is because the VERTICAL component of velocity in both cases is the same.
When a light (the sun) shines on you or any object, the object is in the way of the sunshine hitting the ground on the opposite side of the object, creating shade, or a shadow.
A solid object hitting the ground with the force of a meteor would leave a crater, a depression in the ground with a raised edge at the surface, similar to Meteor Crater in Arizona.
The answer is 91 ft, of course!
4 seconds
I have to make two assumptions:-- The object was not thrown down. Its speed was zerojust as it left the dropper's hand.-- There is no air in this question.With those assumptions . . .H = 1/2 G T2T = sqrt( 2H/G )T = sqrt( 2 x 176.4/9.8 ) = 6 seconds
There is no reason for the object to change.
yes, i just test right now
Some of it is real, not all of it.
Absolutely! If an object has been already traveling for 30 seconds. If you start your watch at 30 seconds, then t = 0 at 30 seconds! The velocity of the object 3 seconds BEFORE you start the watch would then be t = -3 and 3 seconds after you start the watch would be t = 3.
False, provided the drop occurs no sooner than the throw, and the ground is flat .
Interesting question. But when the object is at rest the potential energy of the object is 0, on the surface that is. When it is on a height h it's potential energy increase and when it is dropped from that height all that potential energy gets converted to kinetic energy just before hitting the ground. This extra force comes from this kinetic energy.
The knock of the knuckles of a hand against a door. The noise of an explosion. The noise made by a heavy falling object hitting the ground. The sound of a finger snap.