Refer to the diagram below.
Ignoring air resistance, use gravitational acceleration g = 9.8 m/s².
The pole vaulter drops with an initial vertical speed u = 0.
At impact with the pad, velocity v satisfies:
v² = 2 × (9.8 m/s²) × (4.2 m) = 82.32 (m/s)²
v = 9.037 m/s
As the pad compresses by 0.5 m to bring the vaulter to rest,
let the average acceleration (deceleration) be a m/s². Then:
0 = (9.037 m/s)² + 2 × a × 0.5 m
Solving for a gives:
a = - 82.32 / (2 × 0.5) = -82 m/s²
Thus, the deceleration magnitude is 82 m/s².
Answer:
Explanation:
Provided:
The trolley, with mass M, is allowed to roll freely without friction.
The coefficient of friction between the trolley and mass m is
.
A force F is applied to mass m.
The acceleration of the system is

The frictional force will counterbalance the weight of the block.
The frictional force is 





Answer:
The flow rate of water is (300000kg/s) = (300000l/s)
Explanation:
To compute the volume of moving fluid per second in the channel, we consider the channel's section, the water depth, and the fluid velocity:
Volume flow rate = 15m × 8m × (2.5m/s) = 300 m³/s
To find the mass or liters of water flowing per second, multiply the volume of circulating fluid by the water's density:
Flow rate of water = (300m³/s) × (1000kg/m³) = (300000kg/s) = (300000l/s)
It is important to note that 1kg of water is approximately equivalent to 1 liter.
Answer:
Height (h) = 17 m
Velocity (v) = 18.6 m/s
Explanation: This problem can be solved using kinematic motion equations.
Given Data
Initial velocity (u) = 0
Acceleration (a) = g
Time (t) = 1.9 seconds
First, we calculate the height.

Then, we find the final velocity

The acceleration graph is a linear representation described by y=9.8, as it remains constant:
The velocity graph can be represented by y=9.8x (where y signifies velocity and x indicates time):
The displacement graph can be described as y=4.9x^2 (with x as time and y as displacement):
These graphs apply exclusively from x=0 to x=1.9, so disregard other sections of the graphs.