Question

ball is thrown up with a velocity of 19.6m/s

A.  How long will it take to reach the maximum height? 

B. How high it will go?   ​

Answer 1

A. 2 sec

B. 19.6 m

GIVEN:

Initial velocity of ball, u = 19.6 m/s

TO FIND:

• Time taken to reach the maximum height, t

• Maximum height it reaches , h

FORMULA:

• FIRST EQUATION OF MOTION: v = u + a t

• THIRD EQUATION OF MOTION: v² - u² = 2as

Where,

• a ia acceleration.
• v is final velocity.
• s is displacement.

POINTS TO NOTICE:

• In vertical motion, the acceleration is due to gravity. so a = -g

• Negative sign indicates that the acceleration due to gravity is acting downwards.

• When the ball is thrown upwards, the velocity becomes zero at maximum height.

• v = 0 m/s at maximum height , s = h

So the equations of become,

• v = u - gt
• v² - u² = - 2gh

where g = 9.8 m/s²

SOLUTION :

STEP 1 : TO FIND TIME

v = u - gt

0 = 19.6 - 9.8 t

-19.6 = - 9.8 t

t = 19.6/9.8

Time, t = 2 sec

STEP 2 : TO FIND MAXIMUM HEIGHT

v² - u² = -2gh

0 - (19.6)² = - 2 × 9.8 × h

19.6 × 19.6 = 19.6 h

(19.6 × 19.6) / 19.6 = h

Height, h = 19.6 m

ANSWERS:

• Time taken to reach the maximum height, t = 2 sec

• Maximum height it reaches , h = 19.6 m