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A ball thrown vertically upward reaches at a height 20m at two instants of time t1 and t 2 then find t1 t2?
1). 10 s2
2). 8 s2
3). 4s2
4). 2s2
Correct answer is (3). Can younplease explain the answer?
1). 10 s2
2). 8 s2
3). 4s2
4). 2s2
Correct answer is (3). Can younplease explain the answer?
Most Upvoted Answer
A ball thrown vertically upward reaches at a height 20m at two instant...
Problem:
A ball thrown vertically upward reaches a height of 20m at two instants of time t1 and t2. The options given are:
1) 10 s^2
2) 8 s^2
3) 4 s^2
4) 2 s^2
Solution:
To solve this problem, we need to use the equations of motion for vertical motion under gravity. When an object is thrown vertically upward, it experiences a constant acceleration due to gravity, which is equal to -9.8 m/s^2.
Let's break down the solution into steps:
Step 1: Finding the time taken to reach the maximum height:
When the ball reaches its maximum height, its final velocity becomes zero. We can use the equation:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
In this case, the final velocity is 0 m/s, the initial velocity is the velocity at which the ball is thrown upwards, and the acceleration is -9.8 m/s^2.
0 = u - 9.8t1
u = 9.8t1
Step 2: Finding the time taken to come back down to a height of 20m:
When the ball comes back down and reaches a height of 20m, we can use the equation:
s = ut + (1/2)at^2
where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time.
In this case, the displacement is -20m (negative because the ball is moving downward), the initial velocity is the same as the initial velocity at which the ball was thrown upwards, and the acceleration is -9.8 m/s^2.
-20 = 9.8t2 - (1/2)(9.8)t2^2
Step 3: Solving the equations:
We have two equations:
1) u = 9.8t1
2) -20 = 9.8t2 - (1/2)(9.8)t2^2
By substituting the value of u from equation 1 into equation 2, we get:
-20 = 9.8t2 - (1/2)(9.8)t2^2
This is a quadratic equation. By solving it, we find that t2 = 4s^2.
Substituting this value of t2 into equation 1, we can find t1:
u = 9.8t1
9.8t1 = 9.8 * 4
Therefore, t1 = 4s^2.
Conclusion:
The time taken to reach a height of 20m at two instants, t1 and t2, is 4s^2. Therefore, the correct option is 3) 4s^2.
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A ball thrown vertically upward reaches at a height 20m at two instants of time t1 and t 2 then find t1 t2?1). 10 s22). 8 s23). 4s24). 2s2Correct answer is (3). Can younplease explain the answer?
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