An object is thrown vertically upwards and rises to a height of 10m. Calculate(1) the velocity with which the object was thrown upwards and (2) the time taken by the object to reach the highest point.
Answers
Answered by
869
Let velocity of object is u
we know, at 10m height object will be rest .
so, final velocity = 0
A/C to kinematics equations,
v² = u² + 2aS
0 = u² +2(-g) × 10
0 = u² - 2 × 100
u² = 200
u = 10√2 m/s
Time taken by the object to reach the highest point is t
v = u + at
0 = 10√2 - 10t
t = 10√2/10 sec
t = √2 sec
we know, at 10m height object will be rest .
so, final velocity = 0
A/C to kinematics equations,
v² = u² + 2aS
0 = u² +2(-g) × 10
0 = u² - 2 × 100
u² = 200
u = 10√2 m/s
Time taken by the object to reach the highest point is t
v = u + at
0 = 10√2 - 10t
t = 10√2/10 sec
t = √2 sec
Answered by
263
Answer:
Explanation:
Distance travelled, s = 10 m
Final velocity, v = 0 m s–1
Acceleration due to gravity, g = 9.8 m s–2 Acceleration of the object, a = –9.8 m s–2. (Upward motion)
(i)
v 2 = u2 + 2a s
0=u2 + 2(-9.8)(10)
-u2= -196
u= 14 m/s
(ii) Thus,
v = u + a t
0 = 14 m s–1 – 9.8 m s–2 × t t = 1.43 s.
Initial velocity, u = 14 m s–1, and Time taken, t = 1.43 s.
Similar questions