A body is thrown vertically upwards with a velocity of 40 m/s . Find the maximum height reached by the object also determine the time taken to reach this height . ( g = 10 m/ s2) *
5 sec , 100m
4 sec , 80m
5 sec , 120m
4 sec , 90 sec
Answers
Answer:
- The maximum height reached by the object is 80 m
- The time taken to reach the maximum height is 4 sec
Explanation:
Given :
A body is thrown vertically upwards with a velocity of 40 m/s
To find :
- the maximum height reached by the object
- the time taken to reach maximum height
Solution :
initial velocity, u = 40 m/s
acceleration, a = -g = -10 m/s²
At the maximum height,
velocity, v = 0 m/s
Let the maximum height be h m
From the third equation of motion,
v² - u² = 2as
Substitute the values,
0² - 40² = 2 × (-10) × h
-1600 = -20h
h = 1600/20
h = 80 m
∴ The maximum height reached by the object is 80 m
Time taken to reach the maximum height = t sec
From the second equation of motion,
S = ut + ¹/₂at²
Substitute the values,
80 = 40t + ¹/₂ × (-10) × t²
80 = 40t - 5t²
5t² - 40t + 80 = 0
5(t² - 8t + 16) = 0
t² - 8t + 16 = 0
t² - 4t - 4t + 16 = 0
t(t - 4) - 4(t - 4) = 0
(t - 4) (t - 4) = 0
⇒ t - 4 = 0 ; t = 4
(OR)
From the first equation of motion,
v = u + at
Substitute the values,
0 = 40 + (-10)t
0 = 40 - 10t
10t = 40
t = 40/10
t = 4 sec
∴ The time taken to reach the maximum height is 4 sec
Answer:
t=4s
Explanation:
this is the answer for this question