A ball is thrown vertically upwards From the top of a building of height 29.4 m and with an initial velocity 24.5 m/sec. If the height H of the ball from the ground level is given by H = 29.4 + 24.5t - 4.9t², then find the time taken by the ball to reach the ground
V = Vo + g*Tr = 0.
24.5 -9.8*Tr = 0,
Tr = 2.5 s. = Rise time or time to reach max height(H).
H = 29.4 + 24.5*2.5 - 4.9*2.5^2 = 60 m. Above gnd.
H = 0.5g*Tf^2 = 60.
4.9Tf^2 = 60,
Tf = 3.5 s. = Fall time.
T = Tr + Tf = 2.5 + 3.5 = 6 s. = Time in flight.
To find the time taken by the ball to reach the ground, we need to find the value of t when the height H is equal to zero.
Given the equation H = 29.4 + 24.5t - 4.9t², we set H equal to zero and solve for t:
0 = 29.4 + 24.5t - 4.9t²
Rearranging the equation, we get a quadratic equation in the form: at² + bt + c = 0:
4.9t² - 24.5t - 29.4 = 0
Now, we can solve this quadratic equation using the quadratic formula:
t = (-b ± sqrt(b² - 4ac)) / 2a
Plugging in the values a = 4.9, b = -24.5, and c = -29.4 into the formula, we can calculate t.
t = (-(-24.5) ± sqrt((-24.5)² - 4 * 4.9 * (-29.4))) / (2 * 4.9)
Simplifying further, we have:
t = (24.5 ± 34.515) / 9.8
This gives us two possible values for t:
t₁ = (24.5 + 34.515) / 9.8
t₂ = (24.5 - 34.515) / 9.8
Calculating these values, we have:
t₁ = 5.996 s (approximately)
t₂ = -1.533 s (approximately)
Since time cannot be negative in this context, the negative value t₂ can be ignored.
Therefore, the ball takes approximately 5.996 seconds to reach the ground.