Answer:

Explanation:
Given:
- angle of launch of projectile from horizontal,

- range of projectile,

<u>We have formula for the range of projectile:</u>

putting the respective values

is the velocity with which Tom should jump to land on the other roof.
The impulse given to the ball is equal to the change in its momentum:
J = ∆p = (0.50 kg) (5.6 m/s - 0) = 2.8 kg•m/s
This is also equal to the product of the average force and the time interval ∆t :
J = F(ave) ∆t
so that if F(ave) = 200 N, then
∆t = J / F(ave) = (2.8 kg•m/s) / (200 N) = 0.014 s
The frictional force is given by F = μmg
<span>where μ is the coeficient of friction. </span>
<span>Work done by frictional force = Fd = μmgd </span>
<span>Kinetic energy "lost" = 1/2 mv² </span>
<span>Fd = μmgd = 1/2 mv² </span>
<span>The m's cancel μgd = v² / 2 </span>
<span>d = v² / 2μg </span>
<span>d = 8² / 2(0.41)(9.8) </span>
<span>d = 32 / (0.41)(9.8) </span>
<span>d = 7.96 </span>
<span>Player slides 8 m . </span>
<span>Note. In your other example μ = 0.46 and v = 4 m/s </span>
<span>d = v² / 2μg </span>
<span>= 4² / 2(0.46)(9.8) </span>
<span>= 8 / (0.46)(9.8) </span>
<span>= 1.77 or 1.8 m.
</span>
Hope i Helped :D
I only know P and V and P is pressure and V is volume