3. 02 ball is thrown from the origin with initial speed Vo at an angle with resp

Question

3. 02 ball is thrown from the origin with initial speed Vo at an angle with respect to the horizontal. A person a distance D from the origin runs with speed v to catch the ball Let Vo = 35 m/s, 32°, and D = 110 m. A) Find the maximum height (in m) of the ball above the ground a) 19.19 b) 5.70 17.55 ,d) 16.63 Find the speed of the ball (in m/s) when it passes througn its point.of maximum height. a) 30)) ,b) 35 , 23 , d) 24 C) Find the distance R (in m) to where the balf will return to the ground a) 138.5 b) 48.988.3 dY 112.3 D) Find the time (in s) it would take for the ball to go from the origin to x R. a) 2.16 b) 3.68c) 3.79 d) 3.96 E) How fast v (in m/s) would the person have to run to just be able to catch the b a) 3.18 b) 0.62 c) 4.66 d) 37.63

Explanation / Answer

here,

v0 = 35 m/s

theta = 32 degree

a)

the maximum height , h = (vo * sin(theta))^2 /( 2 * g)

h = ( 35 * sin(35))^2 /( 2 * 9.8) m

h = 17.5 m

b)

the speed of ball at maximum height , vx = v0 * cos(theta) = 30 m/s

c)

the distance , R = (v^2 * sin(2*theta))/g

R = ( 35^2 * sin(2*32)) /9.81

R = 112.3 m

d)

the time of flight , T = 2 * v0 * sin(theta) /g

T = 2 * 35 * sin(32)/9.81 s

T = 3.78 s

e)

the speed , v' = ( D- R)/T

v' = ( 110 - 112.3) /3.78

v' = 0.62 m/s

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