a.)A ball rolls at a constant velocity of 1.25 m/s at an angle of 45 below the +

Question

a.)A ball rolls at a constant velocity of 1.25 m/s at an angle of 45 below the +x-axis in the fourth quadrant.

If we take the ball to be at the origin at t=0 what are its coordinates (x, y) 1.65 s later?

Enter your answers numerically separated by a comma.

b.)Aball with a horizontal speed of 1.8 m/s rolls off a bench 3.7 m high. How long will the ball take to reach the floor? How far from a point on the floor directly below the edge of the bench will the ball land?

c.) An electron is ejected horizontally at a speed of 1.6×106 m/s from the electron gun of a computer monitor. If the viewing screen is 39 cm from the end of the gun,

how far will the electron travel in the vertical direction before hitting the screen?

d.)A ball rolls horizontally with a speed of 8.9 m/s off the edge of a tall platform.

If the ball lands 7.2 m from the point on the ground directly below the edge of the platform, what is the height of the platform?

Express your answer using two significant figures.

Explanation / Answer

a)

here

d = s * t

d = 1.25 * 1.65

d = 2.0625 m

then for the coordinates

sin(-45) = y/2.0625

y = -1.458 m

cos(-45) = x/2.0625

x = 1.458 m

so the coordinates are (1.458,-1.458)

b)

here by using the formula

d = 0.5 * g * t^2

t = sqrt( 2 * d / g)

t = sqrt( 2 * 3.7 / 9.8)

t = 0.8689 sec

the time taken by ball to reach the floor is 0.8689sec

then

the point on the floor is = v * t = 1.8 * 0.8689 = 1.564 m

c)

the time t = d / v

t = 0.39 / (1.6 * 10^6)

t = 2.4375 * 10^-7 sec

then

s = 0.5 * g * t^2

s = 0.5 * 9.8 * (2.4375 * 10^-7)

s = 1.19 * 10^-6 m

d)

x = v * t

t = 7.2 / 8.9

t = 0.8089 sec

then the height of platform is

x = 0.5 * g * t^2

x = 0.5 * (-9.8) * (0.8089)^2

x= -3.2 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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