If I toss an object upward I would expect its upward velocity to decrease by 9.8

Question

If I toss an object upward I would expect its upward velocity to decrease by 9.81 m/s^2 until It readied its maximum altitude Its velocity would be zero there; then its velocity would change sign. Then the magnitude of Its velocity will increase as it heads downward faster and faster If I toss an ob)ect upward at an angle theta_2 to the horizontal its initial y component will be and its Initial horizontal velocity will be As time goes on neglecting air resistance, while where g=9.81 Answers should be to 3 significant figures. Please write your answers so I can read them. Because - signs are so easy to erase. I may exercise the right to make a copy of the examples you hand in. 1 launch a golf ball at 80.0 m/s at an angle of 50.0 degrees above the horizontal. What is its horizontal velocity at t = 2.50 s?. What is the balls vertical (component) of velocity at t=1.30 sec? What is the ball's maximum altitude? At what time after launch is this maximum altitude attained? What horizontal distance wall it have travelled at this point in time? What is y as a function of x for this problem?

Explanation / Answer

initial speed V=80 m/sec

angle, theta=50 degrees,

1)

horizontal velocity,vx=v*cos(theta)

vx=80*cos(50)

vx=51.42 m/sec

2)

vertical velocity, vy=v*sin(theta)-g*t

at t=1.3 sec

vy=80*sin(50)-9.81*1.3

vy=48.53 m/sec

3)

maximum height, H=u^2*sin^2(theta)/(2*g)

H=80^2*sin^2(50)/(2*9.81)

H=191.42 m

4)

time of flight to reach maximum height,

T=u*sin(theta)/g

T=80*sin(50)/9.81

T=6.25 sec

5)

horizontal range is R=(x-xo)=vx*T

R=51.42*6.25

R=321.4 m

here,

y=tan(theta)*x-g/(2*u^2*cos^2(theta))*x^2

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