Question

A objects is thrown vertically up with a speed of 50 m/s. find the maximun hight obtained by the objects​

Answer 1

Given:-

• Initial velocity ,u = 50m/s

• Final Velocity ,v = 0m/s

• Acceleration due to gravity ,a = 9.8m/s

To Find:-

• Maximum height ,h

Solution:-

Using 3rd Equation of motion

• v² = u² +2ah

where

v is the Final velocity

u is the initial velocity

a is the acceleration

h is the height

Substitute the value we get

➨ 0² = 50² + 2×(-9.8) ×h

➨ 0 = 2500 + -19.6 ×h

➨ -2500 = -19.6 × h

➨ h = -2500/-19.6

➨ h = 2500/19.6

➨ h = 127 .55 m

Therefore , the maximum height attained by the object is 127. 55 m.

Additional Information !!

Some equation of motion are

• v = u +at

• s = ut +1/2at².

• v² = u² + 2as

Here,

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement or Distance

t is the time taken

Answer 2

Answer:

$\huge \bf \: given$

• Initial velocity (u) = 50 m/s
• Final velocity (v) = 0 m/s
• Acceleration (A) = 9.8 m/s

$\huge \bf \: to \: find$

Maximum height

$\huge \bf \: solution$

$\huge \bf \: {v}^{2} = {u}^{2} + 2ah$

Putting values

$\sf \implies \: {0}^{2} = {50}^{2} + 2(-9.8)(h)$

$\sf \implies \: 0 = 2500 + - 19.6 \times h$

$\sf \implies \: -2500 = -19.6 \times h$

$\sf \implies \: h = \frac{ - 2500}{-19.6}$

$\sf \implies \: h \: = \frac{2500}{19.6}$

$\sf \implies \: h \: = 127.55m$

$\huge \bf \: height \small \tt \:{maximum} = \huge \bf \: 127.55m$