Question

an object thrown upwards reaches the highest point in 5. 0sec find the velocity with which it was thrown also find the distance covered

Answer 1

Answer:

the initial velocity of the object is 39.2 m/sec.

the distance covered is 78.4 m.

Explanation:

Answer 2

Given:  Time, t = 5 sec

To Find: initial velocity of the object, u.

             distance covered by the object, s.

Solution:

To calculate u, the formula used:

• v - u = a x t ⇒ First law of motion equation
• here, v is the final velocity of the object which is zero at the highest point ( v= 0).
• a is the acceleration, which is g i.e. acceleration due to gravity.
• g = - 9.8m/s²
• Negative sign indicates that the object is thrown against gravity.

Applying the formula for u:

0 - u = -(9.8) x 5

    u = 9.8 x 5

       = 49 m/s

  u = 49 m/s.

To calculate s, the formula used:

s = ut + 1/2 ( at²) ⇒ Second law of motion equation

s = 49x 5 + 1/2 ( 9.8 x 5x5)

 = 245 + 1/2 (9.8 x 25)

 = 245 + 1/2 x 245

 = 245 + 122.5

 = 367.5 m

s = 367.5 m

Hence, the initial velocity of the object is 49 m/s, and the distance covered by it is 367.5 m.