Actually Welcome to the Concept of the Projectile Motion.
Since, here given that, vertical velocity= 50m/s
we know that u*sin(theta) = vertical velocity
so the time taken to reach the maximum height or the time of Ascent is equal to
T = Usin(theta) ÷ g, here g = 9.8 m/s^2
so we get as,
T = 50/9.8
T = 5.10 seconds
thus the time taken to reach max height is 5.10 seconds.
Answer:
patient receiving drug 25 MCG/minute
Explanation:
given data
infusing = 15 ml/hr
drug = 50 mg
D5W = 500 ml
to find out
How many MCG/minute
solution
we know infusing rate is 15 ml/hr = 0.25 ml/min
so 0.25 ml drug content = 50 /500 × 0.25
0.25 ml drug content = 0.025 mg
so here
rate of drug will be 0.025 mg
rate of drug = 0.025 mg = 25 × gm/min
rate of drug = 25 MCG/minute
so patient receiving drug 25 MCG/minute
Answer:
The pitching speed of your friend is 33.20 m/s
Explanation:
<em>Lets explain how to solve the problem</em>
Your friend throw the ball horizontally that means the vertical initial
component of velocity is zero ().
The ball is thrown from a height 4 meters above the ground.
The height
<u><em>Remember:</em></u> the height is negative value because its below the point of
thrown (initial position)
h = -4 m , and g = -9.8 m/s²(downward)
<em>Substitute these values in the rule above</em>
⇒
⇒ -4 = -4.9t² (multiply both sides by -1)
⇒ 4 = 4.9t² (divide both sides by 4.9)
⇒ 0.81633 = t² (take √ for both sides)
⇒ <em>t = 0.9035</em>
Then the time of the ball to land on the ground is 0.9035 seconds
The range of the ball on the ground is 30 m
The range , where
is the horizontal
component of the initial velocity
R = 30 meters and t = 0.9035
⇒ (divide both sides by 0.9035)
⇒ m/s
<em>The pitching speed of your friend is 33.20 m/s </em>