The time taken by the stone to hit the ground would be 5.12 seconds.
<h3>What are the three equations of motion?</h3>
There are three equations of motion given by Newton
The first equation is given as follows
v = u + at
the second equation is given as follows
S = ut + 1/2×a×t²
the third equation is given as follows
v² - u² = 2×a×s
Keep in mind that these calculations only apply to uniform acceleration.
As given in the problem, a stone is dropped from the helicopter which is ascending at the speed of 19.6 m/s
height(S) = 156.8 meters
initial velocity(u) = -19.6 m/s
acceleration(a) = 9.81 m/s²
By using the second equation of motion given by newton
S = ut + 1/2at²
S = 156.8m ,u= -19.6 m/s , a= 9.81 m/s² and t =? seconds
156.8= -19.6t + 9.81t²
t = 5.12 seconds
Thus, the time taken by the stone to hit the ground would be 5.12 seconds.
Learn more about equations of motion from here,
brainly.com/question/5955789
#SPJ1
<span>Ball A with a mass of 0.500 kg is moving east at a velocity of 0.800 m/s. It strikes ball B, also of mass 0.500 kg, which is stationary. Ball A glances off B at an angle of 40.0° north of its...Two smart cars depart from the same starting location at the same time and travel different routes to the same destination, arriving at the same time. Explain why the cars travelled different...Given electric flux density D=0.3r^2 ar nc/m^2 in free space a. Find the total charge within the sphere r=3 b. Find the total electric flux leaving the sphere r=4</span>
Answer:
The mechanical energy of the ball-Earth-floor system the instant the ball left the floor is 7 Joules.
Explanation:
It is given that,
Initial gravitational potential energy of the ball-Earth-floor system is 10 J.
The ball then bounces back up to a height where the gravitational potential energy is 7 J.
Let U is the mechanical energy of the ball-Earth-floor system the instant the ball left the floor. Due to the conservation of energy, the mechanical energy is equal to difference between initial gravitational potential energy and the after bouncing back up to a height.
Initial mechanical energy is 10 + 0 = 10 J
Mechanical energy just before the collision is 0 + 10 = 10 J
Final mechanical energy, 7 + 0 = 7 J
Hence, the mechanical energy of the ball-Earth-floor system the instant the ball left the floor is 7 Joules.
Answer:
The object´s displacement vector is Δr = 8i - 6 j
Explanation:
Hi there!
The position vector is given by the following function:
r = t²i - (3t + 3) j
Let´s find the position of the object at time t1 and t2:
At t1 = 1 s:
r1 = (1)² i - (3 · (1) + 3 )j
r1 = 1 i - 6 j
At t2 = 3 s:
r2 = (3)² i - (3 · (3) + 3) j
r2 = 9 i - 12 j
The displacement is calculated as follows:
displacement = Δr = final position - initial position = r2 - r1
r2 - r1 = 9 i - 12 j - (1 i - 6 j)
r2 - r1 = 9 i - 12 j - 1 i + 6 j
r2 - r1 = 8 i - 6 j
The object´s displacement vector is Δr = 8i - 6 j
I believe the answer is b. external stimulus