Question

a stone is thrown vertically upward with an initial velocity of 40m/s. taking g=10m/s², find the maximum height reached by the stone.what is the net displacement and the total distance covered by the stone?
pls answer breifly​

Answer 1

Answer:

time taken to reach top =4 sec

total time taken in whole journey =8

height /displacement =80 m

total distance = 160m

Explanation:

v=u+at

v=0 at top position

40=10t

t= 4s to go upward

same time will be taken to go downward

v^2 =u^2+2as

40×40=2×10×s

s=80m (displacement)

d=160m(distance

Answer 2

Given :

A stone is thrown vertically upward with an initial velocity of 40 m/s.

To Find :

Maximum height reached by the stone, distance travelled and displacement.

Solution :

❖ For a body thrown vertically upward, g is taken negative as it acts in downward direction.

Since constant acceleration due to gravity acts on the stone, equation of kinematics can be used to solve this question$.$

Third equation of kinematics is given by

• v² - u² = 2gh

» v denotes final velocity

» u denotes initial velocity

» g denotes acceleration

» h denotes height

By substituting the given values

➙ 0² - 40² = 2(-10)h

➙ -1600 = -20h

➙ h = -1600/-20

➙ h = 80 m

∴ Hence distance covered by stone or displacement is 80m.