Solving this using the time, we know that range = horizontal velocity x time of flight
<span>since there are no horizontal forces acting on the ball, there are no horizontal accelerations and the initial horizontal velocity of 36 cos 28 will be constant throughout. I</span><span>f we use the correct time of flight given the launch parameters, we have </span>
<span>range = 36 cos 28 x 3.44 s = 109.3 m </span>
I don’t know sorry ;khbadkhb didhwbck( khwdicdwbihwd
The required spring constant:
The spring constant of the spring is
.
Calculation:
The mass of the car is m=1200 kg, the speed of the car is v=25 m/s, and after colliding the spring is brought to rest at a distance of x=2.5m. Let the spring constant of the spring is, k.
From the conservation of energy,
Total initial kinetic energy= Total final potential energy of the spring
Therefore,

Now, substituting the values of the mass of the car, speed of the car, and displacement, we get:

To know more about spring constant, refer to:
brainly.com/question/14159361
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Answer:
The horizontal distance traveled by the projectile is 15.23 m.
Explanation:
Given;
angle of projection, θ = 25⁰
initial velocity of the projectile, u = 15 m/s
time of flight, t = 1.12 s
The the travelling path of the object is calculated as the range of the projectile

Therefore, the horizontal distance traveled by the projectile is 15.23 m.
I think the answer for question 1 is 4.