Advertisements
Advertisements
प्रश्न
A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s2, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?
Advertisements
उत्तर
Initial velocity (u) = 40m/s
g = −10m/s2
Final velocity at maximum height (v) = 0
Third equation of motion
v2 = u2 + 2gh
0 = 402 + 2 × (−10) × h
1600 = 20h
80m = h
Max Distance = 80m
Total distance travelled by the stone = 80 + 80
= 160m
Because the stone was thrown upwards, but it came back.
Hence, total displacement = 0
APPEARS IN
संबंधित प्रश्न
What is the acceleration of free fall?
A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.
What is the acceleration produced in a freely falling body of mass 10 kg ? (Neglect air resistance)
Give reason for the following :
The force of gravitation between two cricket balls is extremely small but that between a cricket ball and the earth is extremely large.
A pressure of 10 Pa acts on an area of 3.0 m2. What is the force acting on the area ? What force will be exerted by the application of same pressure if the area is made one-third ?
A stone is thrown vertically upwards with initial velocity u reaches a height ‘h’ before coming down. Show that the time taken to go up is the same as the time taken to come down.
Solve the problem.
If the mass of a planet is eight times the mass of the Earth and its radius is twice the radius of the Earth, what will be the escape velocity for that planet?
The value of acceleration due to gravity of earth :
What is the acceleration due to gravity?
Write acceleration due to gravity value on the surface of Earth.
