a coin tossed upward from a balcony 330ft high with an initial velocity of 32/sec. During what interval of time will be the coin be at a height of at least 90ft?
To figure out the time interval during which the coin will be at a height of at least 90 feet, we can use the kinematic equation for vertical motion:
h = h₀ + v₀t - (1/2)gt²
where:
- h is the height of the coin above the ground at any given time t
- h₀ is the initial height of the coin (330 ft in this case)
- v₀ is the initial velocity (32 ft/sec)
- g is the acceleration due to gravity (32 ft/sec²)
To find the time interval when the coin is at least 90 feet high, we solve the equation for time (t):
90 = 330 + 32t - (1/2)(32)t²
Rearranging the equation to get it in standard quadratic form:
16t² - 32t - 240 = 0
Now, we can solve this quadratic equation using the quadratic formula:
t = (-b ± √(b² - 4ac)) / 2a
For this equation, a = 16, b = -32, and c = -240. Plugging in these values:
t = (-(-32) ± √((-32)² - 4(16)(-240))) / (2 * 16)
Simplifying further:
t = (32 ± √(1024 + 15360)) / 32
t = (32 ± √16384) / 32
t = (32 ± 128) / 32
Now, we can find the two time values:
t₁ = (32 + 128) / 32 = 160 / 32 = 5 seconds
t₂ = (32 - 128) / 32 = -96 / 32 = -3 seconds
Since time cannot be negative in this context, we discard the negative value.
Therefore, the coin will be at a height of at least 90 feet for the time interval of 0 seconds to 5 seconds.
it takes a second for the coin to reach max height, 16 ft above the balcony
d = 1/2 a t^2
346 - 90 = 1/2 g t^2
solve for t, add in the 1st second