Name: Weight of an Object on Moon
Uploaded: 2019-07-19
Duration: 7 min 1 s
Description: In many space videos, astronauts are seen hopping on the surface of the moon due to their reduced weights. Watch the video to learn how to find weight on the...

Question

A stone is thrown vertically upward with an initial velocity of

40 m / s

. Taking

g = 10 m / s^{2}

, find the maximum height reached by the stone. What are the net displacement and the total distance covered by the stone?

Accepted Answer

Initial velocity, $u = 40 m / s$
Acceleration due to gravity, $g = - 10 m / s^{2}$ (If an object is thrown vertically up with an initial velocity $u$ , there will be retardation, i.e. $a = — g$ )
Maximum height, $h = S = ?$
Final velocity $v = 0$
From laws of kinematics,
$v^{2} - u^{2} = 2 g S$
$(0)^{2} - (40)^{2} = 2 \times (- 10) \times S$
$S = \frac{- (40 \times 40)}{2 \times (- 10)}$
$S = 80 m$
Net displacement of the stone is equal to zero as the stone falls back to the same point.
Total distance covered by stone $= 80 m + 80 m = 160 m$ .

A stone is thrown vertically upward with an initial velocity of $40 m / s$ . Taking $g = 10 m / s^{2}$ , find the maximum height reached by the stone. What are the net displacement and the total distance covered by the stone?

Important Questions on Gravitation

Learn and Prepare for any exam you want !

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 are the net displacement and the total distance covered by the stone?

Important Questions on Gravitation

Learn and Prepare for any exam you want !

A stone is thrown vertically upward with an initial velocity of $40 m / s$ . Taking $g = 10 m / s^{2}$ , find the maximum height reached by the stone. What are the net displacement and the total distance covered by the stone?