Question

after how much time a body will strike to the ground if it through with upwards velocity 20m/s.​

Answer 1

Answer:

Method I: We will breah the motion in two parts.

O to A (vertically upward motion):

At the highest point, velocity=0

v=u−gt

0=20−10t1⇒t1=2s

v2=u2−2gh

0=(20)2−2×10×h⇒h=20m

A to B (vertically downward direction):

AB=h+60=20+60=80m

80=0+12gt22=5t22⇒t2=4s

The ball will strike the ground after 6s.

v0=u+gt2=0+10×4=40m/s

Note: The above method is lengthy. We will not break motion in part unless necessary. We know that equations of motion are applicable for displacement.

Method II: Now taking O as the origin and vertically upward direction positive, from O to B through A, the displacement is negative, i.e.

h=−60m

h=ut−12gt2(u↑⏐g⏐↓)

−60≡20t−5t2

t2−4t−12=0

(t−6)(t+2)=0

t=6s,t=−2s

Time t=−2s is not possible, time cannot be-ve.

The ball will steike the ground after 6s.

v=u−gt

v0=20−10×6=−40m/s

Negative sign indicates that the velocity is in downward direction.

Answer 2

Answer :-

4s

Explanation :-

Given :

A object is thrown upward with,

Initial velocity, u = 20m/s

Final velocity,v = 0 m/s [At maximum height, final velocity of the object become zero, i.e., v= 0]

To Find :

After how much time a body will strike to the ground,t = ?

Solution :

During upward motion, g= -10m\s²

According the first equation of motion,

$\sf{}v=u+at$

$\sf{}\therefore v=u-gt$

Put their values and find “t”

$\sf{}\implies 0=20-10\times t$

$\sf{}\implies -20=-10\times t$

$\sf{}\implies \dfrac{-20}{-10}=t$

$\sf{}\therefore t=2s$

Time taken to reach the ground is 2s + 2s = 4s seconds. [As the object will come back to the ground after throwing it up with the same amount of time]