Answer:
[tex]v_{2}=28.205m/s[/tex]
Explanation:
Given data
Initial altitude of ball is y₁=27.5m and its initial velocity is v₁=16.0 m/s
We know that,if other forces than gravitational do work,the total work done by all forces is given by:
[tex]W_{tot}=W_{grav}+W_{other}=K_{2}-K_{1}[/tex]
And since the work done is by gravitational force is given by:
Wgrav= -ΔUgrav =mgy₁-mgy₂
And since the kinetic energy is given by:
[tex]K=1/2mv^2[/tex]
So we get
[tex]\frac{1}{2}mv_{1}^2+mgy_{1}+W_{other}=\frac{1}{2}mv_{2}^2+mgy_{2}[/tex]
When the ball is thrown there is no other force than the gravitational force which acts on the ball So we have
[tex]W_{other}=0[/tex]
Taking the ground to be zero potential energy y₂=0
So we get
[tex]\frac{1}{2}m(16.om/s)^2+m(9.81m/s^2)(27.5m)+0=\frac{1}{2}mv_{2}^2+0\\ 397.775=1/2v_{2}^2\\v_{2}=\sqrt{2*397.775} \\v_{2}=28.205m/s[/tex]
Given Information:
distance = y = 27.5 m
Angle = θ = 37°
initial speed of the ball = v1 = 16 m/s
Required Information:
final speed of the ball = v2 = ?
Answer:
v2 = 28.19 m/s
Explanation:
We will write an equation in terms work done by the forces acting on the ball. We know that one force is gravitational force
W = mgy1 - mgy2
kinetic energy is given by
KE = 0.5mv1² so
0.5mv1² + mgy1 = 0.5mv2² + mgy2
Mass cancels out and y2 is zero so equation becomes
0.5v1² + gy1 = 0.5v2²
0.5v2² = 0.5v1² + gy1
v2² = 2*(0.5v1² + gy1)
v2 = √2*(0.5v1² + gy1)
Substitute the values
v2 = √2*(0.5(16)² + 9.8*27.5)
v2 = 28.19 m/s
Therefore, the speed of the ball is 28.19 m/s just before it strikes the ground.
