A ball is thrown upward. Its initial vertical component of velosity is 30m/s and its initial horizontal component of velosity is 20m/s. What is the ball's speed 2 s later? Give: Viy = 30m/s Vix = 20m/s T = 2 sec
a) V = gt =10m/sec2 x 2 sec= 20 m/sec
b) c2 = (20m/s)2 + (20m/s)2 = 400m2/s2 +400m2/2 = 800m2/s2 c = 28.28 =Vf V =gt = 10m/s2(2 sec) = 5m/s
Xo = 20 m/s
Yo = 30 m/s
Vo = sqrt(20^2+30^2) = 36.1 m/s
Y = Yo + g*t = 30 + (-9.8)*2 = 10.4 m/s.
= Ver. component of velocity 2 s later.
X = Xo = 20 m/s and remains constant.
V^2 = X^2 + Y^2 = 20^2 + 10.4^2 = 508.16
V = 22.5 m/s. = Total velocity.
c) The ball's speed 2 seconds later is 28.28 m/s, as calculated using the Pythagorean theorem with the horizontal and vertical components of velocity. But be careful, this ball might be moving faster than a clown running away from an angry mob!
To find the ball's speed 2 seconds later, we need to calculate the magnitude of its velocity. The velocity can be represented as the sum of its vertical and horizontal components.
Given:
Viy = 30 m/s (vertical component of velocity)
Vix = 20 m/s (horizontal component of velocity)
T = 2 seconds (time)
We can calculate the vertical component of the velocity using the equation: Viy = gt.
Viy = 10 m/s^2 * 2 s = 20 m/s.
Next, we can calculate the horizontal component of the velocity using the equation: Vix = V (final velocity) - g * T.
20 m/s = V - 10 m/s^2 * 2 s.
20 m/s = V - 20 m/s^2.
V = 40 m/s.
Now, we can find the magnitude of the velocity using the Pythagorean theorem, where c^2 = (Vi)^2 + (Vj)^2.
c^2 = (20 m/s)^2 + (30 m/s)^2.
c^2 = 400 m^2/s^2 + 900 m^2/s^2.
c^2 = 1300 m^2/s^2.
c ≈ 36.06 m/s (rounded to two decimal places).
Therefore, the ball's speed 2 seconds later is approximately 36.06 m/s.
To find the ball's speed 2 seconds later, we need to calculate the magnitude of its velocity at that time.
Given:
Initial vertical component of velocity (Viy) = 30 m/s
Initial horizontal component of velocity (Vix) = 20 m/s
Time (T) = 2 seconds
To find the ball's speed 2 seconds later, we can use the Pythagorean theorem:
c^2 = a^2 + b^2
Where:
c is the magnitude of the velocity (speed)
a is the vertical component of velocity (Viy)
b is the horizontal component of velocity (Vix)
Let's solve for the magnitude of the velocity:
c^2 = (Viy)^2 + (Vix)^2
c^2 = (30 m/s)^2 + (20 m/s)^2
c^2 = 900 m^2/s^2 + 400 m^2/s^2
c^2 = 1300 m^2/s^2
Taking the square root of both sides to find c:
c = √(1300 m^2/s^2)
c ≈ 36.06 m/s
Therefore, the ball's speed 2 seconds later is approximately 36.06 m/s.