A particle is projected vertically upwards from a point O on the ground. It takes time t_(1) to reach a point A at a height h above the ground, it continues to move and takes a time t_(2) to reach the ground. Find (a) h, (b) the maximum height reached and (c ) the velocity of the partical at the half of maximum height.
A particle is parojected vertically upwards from grund with initial velocity u . a. Find the maximum height H the particle will attain and time T that it will attain and time T that it will take to return to the ground . . b. What is the velocity when the particle returns to the ground? c. What is the displacement and distance travelled by the particle during this time of whole motion.
A particle is projected vertically upwards from ground with velocity 10 m // s. Find the time taken by it to reach at the highest point ?
A particle is projected vertically upwards with a velocity v. It returns to the ground in time T. Which of the following graphs correctly represents the motion.
A particle is projected vertically upward from the ground at time t = 0 and reaches a height h at t = T . Show that the greater height of the particle is (g T^(2) + 2h)^(2)//8 gT^(2)
A particle is projected vertically upwards from a point A on the ground. It takes t_(1) time to reach a point B but it still continues to move up. If it takes further t_(2) time to reach the ground from point B then height of point B from the ground is :-
A particlele is projected vertically upwards from the ground with a speed V and a speed v and a second particle is projected at the same instant from a height h directly above the first particel with the same speed v at an angle of projection theta with the horizontal in upwards direction. The time when the distance between them is minimum is
A particle is thrown vertically upward from the ground with some velocity and it strikes the ground again in time 2 s. The maximum height achieved by the particle is : (g=10 m//s^(2))
A particle is projected vertically upwards from ground with an initial speed of 57m/s .If acceleration due to gravity is taken to be equal to 10m/ s^(2) ,then the displacement of the particle in the sixth second of its motion will be?
A particle when projected vertically upwards from the ground, takes time T to reach maximum height H. Find out (i) The height of the particle from ground at T//2,2T//3,4T//3,5T//4 and also find the ratio of K.E. to P.E. for particle at the same instant of time. (ii) If the particle crosses a point at height 7H//16 at time t_(1) and t_(2) then find t_(1)//t_(2)
_2019_CBT3_E01_050_O01.png)
_2019_CBT3_E01_050_O02.png)
_2019_CBT3_E01_050_O03.png)
_2019_CBT3_E01_050_O04.png)
