Answer:
The time taken by the rock to reach the ground is 0.569 seconds.
Explanation:
Given that,
A student throws a rock horizontally off a 5.0 m tall building, s = 5 m
The initial speed of the rock, u = 6 m/s
We need to find the time taken by the rock to reach the ground. Using second equation of motion to find it. We get :
So, the time taken by the rock to reach the ground is 0.569 seconds. Hence, this is the required solution.
Answer:
terminal velocity is;
v = 117.54 m/s
v = 423.144 km/hr
Explanation:
Given the data in the question;
we know that, the force on a body due to gravity is;
= mg
where m is mass and g is acceleration due to gravity
Force of drag is;
=
pCAv²
where p is the density of fluid, C is the drag coefficient, A is the area and v is the terminal velocity.
Terminal velocity is reach when the force of gravity is equal to the force of drag.
mg = pCAv²
we solve for v
v = √( 2mg / pCA )
so we substitute in our values
v = √( [2×(86 kg)×9.8 m/s² ] / [ 1.21 kg/m³ × 0.7 × 0.145 m²] )
v = √( 1685.6 / 0.122015 )
v = √( 13814.6949 )
v = 117.54 m/s
v = ( 117.54 m/s × 3.6 ) = 423.144 km/hr
Therefore terminal velocity is;
v = 117.54 m/s
v = 423.144 km/hr
Answer:
Eleven seconds.
Explanation:
Two keys are needed to solve this problem. First, the conservation of momentum: allowing you to calculate the cart's speed after the elephant jumped onto it. It holds that:
So, once loaded with an elephant, the cart was moving with a speed of 4.29m/s.
The second key is the kinematic equation for accelerated motion. There is one force acting on the cart, namely friction. The friction acts in the opposite direction to the horizontal direction of the velocity v0, its magnitude and the corresponding deceleration are:
The kinematic equation describing the decelerated motion is:
It takes 11 seconds for the comical elephant-cart system to come to a halt.