An object is thrown vertically upward and reaches to 98m then starts falling freely. Find the time of flight of the body and the velocity through which it was thrown in km/h.
Answers
Explanation:
Given Initial velocity of ball, u=49 m/s
Let the maximum height reached and time taken to reach that height be H and t respectively.
Assumption: g=9.8 m/s
2
holds true (maximum height reached is small compared to the radius of earth)
Velocity of the ball at maximum height is zero, v=0
v
2
−u
2
=2aH
0−(49)
2
=2×(−9.8)×H
⟹H=122.5 m
v=u+at
0=49−9.8t
⟹t=5 s
∴ Total time taken by ball to return to the surface, T=2t=10 s
Answer:
The time of flight of the body is 9 s.
The velocity through which it was thrown is 157.7 Km/hr.
Explanation:
An object is thrown vertically upward and reaches 98 m then starts falling freely.
Final velocity, v = 0 m/s
Distance covered, S = 98 m
Acceleration, a = g = -9.8 m/s²
Since the motion is against gravity the sign of acceleration due to gravity will be negative.
Initial velocity can be calculated using the 3rd equation of motion.
v² = u² + 2aS
u² = v² - 2aS
u² = 0² - (2 × -9.8 × 98)
u² = 1920.8
u = √1920.8
u = 43.8 m/s
To convert this value of initial velocity form m/s to Km/hr multiply the value with 18/5.
∴ Initial velocity, u = (43.8 × 18) / 5
u = 157.7 Km/hr
The time taken to cover the distance can be calculated using the 1st equation of motion.
v = u + at
t = (v - u) / a
t = (0 - 43.8) / -9.8
t = 4.5 s
Time taken to reach the height is 4.5 seconds. The object has to travel up and down. So the total time of flight will be:
t = 4.5 + 4.5
t = 9 s