Question

An object is thrown vertically upwards and rises to a height of 10 m. Calculate (i) the velocity with which the object was thrown upwards and (ii) the time taken by the object to reach the highest point.

Answer 1

Answer :-

(i) Initial velocity, u = 14 m s

(ii) Time taken, t = 1.43 s.

Explanation :-

Given :

Distance traveled, s = 10m

Final velocity, v = 0m/s

Acceleration due to gravity, g = -9.8m/s^2

To Find :

i) Initial velocity  of the ball,u = ?

ii) Time taken,t = ?

Solution :

According to the third equation of motion,

$\boxed{\sf{}v^2=u^2+2as}$

Put their values and solve it,

$\sf{}:\implies 0=u^2+2\times -9.8\times 10$

$\sf{}:\implies -u^2=2\times -9.8\times 10$

$\sf{}:\implies -u^2=-196$

$\sf{}:\implies u^2=196$

$\sf{}:\implies u=\sqrt{196}$

$\sf{}\therefore u=14$

Therefore,initial velocity is equal to 14m/s

According to the first equation of motion,

$\boxed{\sf{}v=u+at}$

Pu their values and solve it,

$\sf{}:\implies \sf{}0=14+(-9.8)\times t$

$\sf{}:\implies -14=-9.8\times t$

$\sf{}:\implies \dfrac{-14}{-9.8}=t$

$\sf{}:\implies t=1.42s$

Therefore,time is equal to 1.42s

Answer 2

(i) Initial velocity, u = 14 m s

(ii) Time taken, t = 1.43 s.

Given :

Distance traveled, s = 10m

Final velocity, v = 0m/s

Acceleration due to gravity, g = -9.8m/s²

Solution :

According to the third equation of motion,

:⟹0 = u 2 +2×−9.8×10

$\sf{}:\implies -u^2=2\times -9.8\times 10$

$\sf{}:\implies -u^2=-196$

$\sf{}:\implies u^2=196$

$\sf{}:\implies u=\sqrt{196}$

$\sf{}\therefore u=14$

Therefore,initial velocity is equal to 14m/s

⭐According to the first equation of motion,⭐

$\star \red{v=u+at}$

⟹0=14+(−9.8)×t

$\sf{}:\implies -14=-9.8\times t:$

$\sf{}:\implies \dfrac{-14}{-9.8}=t$

$\sf{}:\implies t=1.42s$

Therefore,time is equal to 1.42s

✡more info✡

• the SI unit of pressure is Pascal.
• Pascal is written as (Pa).
• the SI unit of force is Newton.
• Newton is written as (N).
• 1 kgf = 9.8 N m
• 1 N m = 10⁷ dyne cm