Answer:
t = √2y/g
Explanation:
This is a projectile launch exercise
a) The vertical velocity in the initial instants (
= 0) zero, so let's use the equation
y = t -1/2 g t²
y= - ½ g t²
t = √2y/g
b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity
x = vox t
x = v₀ₓ √2y/g
c) Speeds before touching the ground
vₓ = vox = constant
= - gt
= 0 - g √2y/g
= - √2gy
tan θ = Vy / vx
θ = tan⁻¹ (vy / vx)
θ = tan⁻¹ (√2gy / vox)
d) The projectile is higher than the cliff because it is a horizontal launch
Answer:
Explanation:
Given data:
Mass of the paper clip, 
Kinetic energy, 
Let the velocity of the paper clip when it is thrown be <em>v</em>.
Thus,
(rounding to nearest tenth)
Urbfhdhehjrrr
Rrrrdndndndjdjdjdjjdjdjdjdkdmdmdmdkdjdjdjdjdid
Answer:
The right approach is Option b (the force..................exert on you).
Explanation:
- Even before you fall on something like a soft object, users eventually slow to a halt. You are still giving away all the downward momentum, but progressively although with small powers, you are doing so.
- Although you can get injured by massive powers, this gradual displacement is a positive thing. And that is why you have a mattress you would like to settle on.
The other options given are not connected to the situation described. So, the solution here was the right one.
Answer:
Speed of another player, v₂ = 1.47 m/s
Explanation:
It is given that,
Mass of football player, m₁ = 88 kg
Speed of player, v₁ = 2 m/s
Mass of player of opposing team, m₂ = 120 kg
The players stick together and are at rest after the collision. It shows an example of inelastic collision. Using the conservation of linear momentum as :
V is the final velocity after collision. Here, V = 0 as both players comes to rest after collision.
So, the speed of another player is 1.47 m/s. Hence, this is the required solution.
