A ball is thrown at a speed of 20.0 m/s at an angle of 30.0o above the horizonta
ID: 2056632 • Letter: A
Question
A ball is thrown at a speed of 20.0 m/s at an angle of 30.0o above the horizontal. a. What is the horizontal component of the velocity? b. What is the vertical component of the velocity? c. What is the horizontal component of the velocity 2.00s after being thrown? d. What is the vertical component of the velocity 2.00s after being thrown? e. How long did the ball take to reach its highest point? f. What is the time of the flight for the projectile? g. How high did the ball rise? h. What is the range of the ball?
please show all work, im having hard time understanding this, so please help and show all work, step by step...thanks!!!!!!!!!!!
A ball is thrown at a speed of 20.0 m/s at an angle of 30.0o above the horizontal. a. What is the horizontal component of the velocity? b. What is the vertical component of the velocity? c. What is the horizontal component of the velocity 2.00s after being thrown? d. What is the vertical component of the velocity 2.00s after being thrown? e. How long did the ball take to reach its highest point? f. What is the time of the flight for the projectile? g. How high did the ball rise? h. What is the range of the ball?Explanation / Answer
u = 20m/s =30
a) horizontal comp, ux =20cos = 17.32 m/s
b) vertical comp, uy = 20sin = 10m/s
c) ux = constant as ax= 0 ( no acc. in horizontal dirn)
ux after t=2sec = 17.32 m/s
d) uy = uyini + at = 10 - 9.8*2 = -9.6m/s
e) uy =0
uyini - 9.8t = 0
t = 10/9.8 =1.02 sec
f) time of flight = 2*(time taken to reach highest point)
= 2*1.02 =2.04 sec
g) max height , h = uyini2/2g = 100/2*9.8 =5.1 m
h) range = ux*time of flight = 17.3*2.04 =35.29 m
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