A ball is thrown vertically upward with a speed of 1.53 m/s from a point 4.21 m above the ground. Calculate the time in which the ball reach the ground.
I'm figuring I have to do something with 9.80 which is gravity but I don't know what.
heightfinal=heightinitial + Vi*time-1/2 g time^2
solve for time. Use the quadratic equation.
Calculate the time T1 it takes to rise to maximum altitude H. Velocity will be zero there. This time can be obtained from
g T1 = 1.53 m/s
Calculate the height at that time from
H = 4.21 + 1.53 T1 - (1/2) g T1^2
Then calculate the time T2 it takes to fall to the ground from that height H. Get that by solving
H = (1/2) g T2^2
The total time of flight is T1 + T2
I'm alittle confused.
on the very first part do I multiply 9.80*1.53 or is the time I need 1.53
You are confused. In my equation, you solve for time. It is a quadratic. DrWls gave an alternate approach, but you are still solving for time. In his approach, the total time is the sum of the times to get up to max altitude, and the time to fall to the ground.
To calculate the time it takes for the ball to reach the ground, you can use the kinematic equation that relates displacement, initial velocity, time, and acceleration:
s = ut + (1/2)at^2
Where:
s = displacement (in this case, the height of the ball, which is -4.21 m since it is moving downwards)
u = initial velocity (1.53 m/s, upwards)
t = time
a = acceleration (in this case, acceleration due to gravity, which is approximately -9.8 m/s^2, downwards)
Substituting the given values into the equation, we have:
-4.21 = (1.53)t + (1/2)(-9.8)(t^2)
Now, we can solve this equation to find the value of t. Rearranging the equation, we get:
(1/2)(-9.8)t^2 + (1.53)t - 4.21 = 0
We can solve this quadratic equation either by factoring, completing the square, or using the quadratic formula. Using the quadratic formula is the most common and straightforward method. The quadratic formula is:
t = (-b ± √(b^2 - 4ac)) / (2a)
Here, a = (1/2)(-9.8), b = 1.53, and c = -4.21. Plugging these values into the quadratic formula, we get:
t = [-(1.53) ± √((1.53)^2 - 4 * (1/2)(-9.8)(-4.21))] / (2 * (1/2)(-9.8))
Simplifying this expression further, we get:
t = [-(1.53) ± √(2.3509 + 41.4402)] / (-9.8)
Now, calculate the values inside the square root and further simplify the expression:
t = [-(1.53) ± √(43.7911)] / (-9.8),
t = [-(1.53) ± 6.6159] / (-9.8).
Now, we'll have two possible values for t:
t1 = (-(1.53) - 6.6159) / (-9.8),
t2 = (-(1.53) + 6.6159) / (-9.8).
Calculating t1 and t2, we get:
t1 ≈ 0.76 seconds,
t2 ≈ -0.63 seconds.
Since time cannot be negative, the ball reaches the ground after approximately 0.76 seconds.