3. You throw a ball, m= 2kg, with an initial velocity of 25 m/s at an angle of 5

Question

3. You throw a ball, m= 2kg, with an initial velocity of 25 m/s at an angle of 50 degrees from the top of a 30m cliff.
a. Using energy considerations, what is the maximum height of the ball?
b. What is the magnitude of the velocity when it hits the ground?
c. Now, using KINEMATICS, what is the magnitude of the velocity when it hits the ground?

4. You compress a spring with k = 3000 N/m a distance of 0.2 m with a block of mass 4 kg. You then release the block. It slides along a frictionless surface and then goes up an incline of 35 degrees.
a. What is the velocity of the block after it leaves the spring but before it gets to the incline?
b. How far up the incline (along the hypotenuse of the triangle) does the block travel?

Explanation / Answer

here,

3)

mass , m = 2 kg

initial velocity , u = 25 m/s

theta = 50 degree

a)

the speed of ball at maximum height , v1 = u * cos(theta)

let the maximum height be hmax

using conservation of energy

0.5 * m * u^2 + m * g * h = 0.5 * m * (u * cos(theta))^2 + m *g * hmax

0.5 * 25^2 + 9.81 * 30 = 0.5 * ( 25 * cos(50))^2 + 9.81 * hmax

solving for hmax

hmax = 48.7 m

the maximum height is 48.7 m

b)

let the final speed be v

using conservation of energy

0.5 * m * u^2 + m * g * h = 0.5 * m * v^2

0.5 * 25^2 + 9.81 * 30 = 0.5 * v^2

solving for v

v = 34.8 m/s

the magnitude of the velocity is

c)

let the final vertical velocity be vy

vy^2 - (u * sin(theta))^2 = 2 * g * h

vy^2 - (25 * sin(50))^2 = 2 * 9.81 * 30

solving for vy

vy = 30.9 m/s

vx = u * cos(theta) =

the magnitude of final velocity , v = sqrt(vy^2 + vx^2)

v = 34.8 m/s

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.