A particle is projected up with an initial velocity of 80 ft /sec. The ball will be at a height of 96 ft from the ground after :- 2.0 and 3.0 secOnly at 3.0 secOnly at 2.0 secAfter 1 and 2 sec
Question
A particle is projected up with an initial velocity of 80 ft /sec. The ball will be at a height of 96 ft from the ground after :- 2.0 and 3.0 secOnly at 3.0 secOnly at 2.0 secAfter 1 and 2 sec
Solution
To find out when the particle will be at a height of 96 ft, we can use the equation of motion for vertical motion:
h = h0 + v0t - (1/2)gt^2
where: h is the height of the particle at time t h0 is the initial height (in this case, 0 ft) v0 is the initial velocity (in this case, 80 ft/sec) g is the acceleration due to gravity (approximately -32 ft/sec^2) t is the time in seconds
Let's calculate the time it takes for the particle to reach a height of 96 ft:
96 = 0 + 80t - (1/2)(-32)t^2
Simplifying the equation, we get:
32t^2 - 80t + 96 = 0
Now we can solve this quadratic equation to find the values of t when the particle is at a height of 96 ft.
Using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
where a = 32, b = -80, and c = 96, we can substitute these values into the formula:
t = (-(-80) ± √((-80)^2 - 4(32)(96))) / (2(32))
Simplifying further:
t = (80 ± √(6400 - 12288)) / 64
t = (80 ± √(-5888)) / 64
Since the discriminant (√(-5888)) is negative, there are no real solutions for t. Therefore, the particle will not be at a height of 96 ft at any time.
Therefore, the correct answer is: None of the given options (2.0 and 3.0 sec, only at 3.0 sec, only at 2.0 sec, after 1 and 2 sec).
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