A ball is thrown with a velocity of 20 m/s; at an of 35 degrees; with a release

Question

A ball is thrown with a velocity of 20 m/s; at an of 35 degrees; with a release height of 2m. Ignoring wind resistance, compute the following: A. Vertical velocity at takeoff B. Horizontal velocity at takeoff C. The balls horizontal acceleration D. Time in flight E. Peak height F. How far will it go in the horizontal direction A ball is thrown with a velocity of 20 m/s; at an of 35 degrees; with a release height of 2m. Ignoring wind resistance, compute the following: A. Vertical velocity at takeoff B. Horizontal velocity at takeoff C. The balls horizontal acceleration D. Time in flight E. Peak height F. How far will it go in the horizontal direction A. Vertical velocity at takeoff B. Horizontal velocity at takeoff C. The balls horizontal acceleration D. Time in flight E. Peak height F. How far will it go in the horizontal direction

Explanation / Answer

here,

initial velocity , u = 20 m/s

angle theta = 35 degree

height of takeoff , h = 2 m

A.

vertical velocity at takeoff , uy = u * sin(theta)

uy = 20 * sin(35)

uy = 11.47 m/s

B.

horizontal velocity at takeoff , ux = u * cos(theta)

ux = 20 * cos(35)

ux = 16.38 m/s

C.

the balls horizontal accelration is zero

D.

let the time of flight be t

-h = uy *t - 0.5 * g*t^2

- 2 = 11.47 * t - 4.9 * t^2

t = 2.5 s

the time of flight is 2.5

E.

let the peak height be H

vy^2 - uy^2 = - 2*g * (H-h)

0 - 11.47^2 = - 2 * 9.8 * (H - 2)

H = 8.71 m

the peak height is 8.71 m from the ground

F.

the distnace travelled in the horizontal direction , d = ux *t

d = 16.38 * 2.5

d = 40.95 m

the distnace travelled in the horizontal direction is 40.95 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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