Answer:
d_total = 12 m
Explanation:
In this kinematics scenario illustrated in the graph provided, we determine the distance traveled over a 24-second duration.
The comprehensive distance can be calculated as follows:
d_total = d₁ + d₂ + d₃
Given that d₂ on the graph is level (v=0), its distance equates to zero, hence d₂ = 0.
The distance for d₁ is calculated as:
d₁ = 12 - 6 = 6 m
For distance d₃:
d₃ = 6 - 0 = 6 m
Thus, the overall distance covered is:
d_total = 6 + 0 + 6
d_total = 12 m
The formula for range is:

Given values are:

where θ equals 14.1 degrees

Using the equation above,

The calculated range is 66.7 meters.
Therefore, the range is approximately 66.1 meters.
A 40 kg child throws a 0.5 kg stone at a velocity of 5 m/s. To find the recoil, we apply the conservation of momentum formula: m1•v1 + m2•v2 = 0, where m1 is the mass of the child, and v1 is the child's recoil velocity. Applying the known values results in 40•v1 = -0.5 × 5, leading to v1 = -2.5 / 40, which simplifies to v1 = -0.0625 m/s. Thus, the child's recoil speed is 0.0625 m/s.
In this scenario, there exists a constant electric field produced by a large sheet. This electric field can be defined as... The force acting on the ball due to this field acts horizontally, and this force must be counterbalanced by the horizontal tension component of the string to maintain equilibrium. Similarly, the vertical tension component in the string must equal the weight of the small sphere. Hence, we can derive two equations to illustrate this.
Answer:
A. Yes, the ball clears the crossbar by 2.83 meters
Explanation:
This situation pertains to projectile motion.
The horizontal velocity component of the ball is calculated as 26 cos 35 = 21.3 m/s
The vertical velocity component is 26 sin 35 = 14.9 m/s
The time taken to travel the horizontal distance to the goalpost, which is 54.9 m, is:
= distance / horizontal speed
= 54.9 / 21.3
= 2.577 seconds.
The vertical distance achieved during this time is:
h = ut - 1/2 gt², where u is the initial vertical velocity, and t = 2.577 seconds.
h = 14.9 x 2.577 - 0.5 x 9.8 x (2.577)²
= 38.39 - 32.54
= 5.85 m
Thus, the ball surpasses the crossbar by 5.85 - 3.05 = 2.8 m