3. Practicing for a golftournarnent Mr. Dormer hits a golf ball is hit or*187 cl

Question

3. Practicing for a golftournarnent Mr. Dormer hits a golf ball is hit or*187 clift. The initiad velocity of the ball is 123 nsecatanangle of 54 low f.hom the base of the cliff does the ball land, and how high is the golf ball off the ground a its highest point? (10 points) high 4. In the final seconds of the game, Green calls for a Hail Mary Pass. The ball is thrown with a velocity of 684ft/sec at an angle of 46.0 The receiver catches the ball on the goal line at the same height it was theown. How far was the touchdown, and how high was the ball at its highest point? (1o points)

Explanation / Answer

3) Let the point from which the ball is hit is O, the highest point be A, and the point where the ball hits the ground is B.

v=37.5 m/s, height of cliff, h1=57 m

distance from O to A

u=vsin(theta)=37.5*sin(54)=30.33 m/s,

final velocity at A is zero(highest point), a=g=9.8m/s2

distance from O to A be h2 (say)

v2-u2=2*a*h2

h2=46.95 m

Time taken from O to A

using, v=u+at1

t1= 30.33/9.8=3.094 seconds

Total height, h= h1+h2=57+46.95= 103.95 m

this is the vertical height from A to B

Time taken from A to B

h=103.95 m, a=g=9.8 m/s2, u=0,

h=ut2+0.5*g*t22

t2=4.065 seconds

Total time taken O to B is =t1+t2= 3.094+4.605=7.699 seconds

Now horizontal journey from O to A

velocity= v*cos(theta)=22.041 m/s, time= 7.699 seconds

Distance(horizontal Range)= velocity*time=7.699*22.041

Range=169.69 m

Total height = 103.95 m

4) this is a question of horizontal projectile

given, v=20.84 m/s, theta=46 degrees, g=9.8 m/s2

Range= v2*sin(2*theta)/g = 20.842* sin(92)/9.8

Range= 44.2899 m

max height= v2*sin2(theta)/2g

Max height=11.46 m

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