Question

a body is dropped freely under gravity from the top of a tower of height 78.4m, calculate time to reach the ground and the velocity with which it strikes the ground,
take g =9.8m.s^2

Answer 1

Given

• Initial Velocity (u) = 0m/s
• Height (h) = 78.4m
• Acceleration due to gravity (g)  = 9.8m/s²

To find

• Time taken (t)
• Final Velocity (v)

Solution

• From 2nd equation of motion:
• h = ut +½ gt²
• 78.4 = 0×t + ½ × 9.8 × t²
• 78.4 = 4.9 t²
• t² = 78.4/4.9 = 16
• t = √16
• t = 4

☯ Time taken by the body to reach the ground = 4s

• From 1st equation of motion:
• v = u + at
• v = 0 + 9.8×4
• v = 39.2

☯ Final velocity or the body strikes the ground with velocity = 39.2m/s²

Learn more

☯ All the three equations of motion:

• v = u + at
• s = ut + ½ at²
• v² - u² = 2as

☯ Where :

• v is final velocity
• u is initial velocity
• a is acceleration
• s is distance
• t is time taken