A projectile is fired in such a way that its horizontal range is equal to three

Question

A projectile is fired in such a way that its horizontal range is equal to three times its maximum height. What is the angle of projection? A fish swimming in a horizontal plane has velocity v = (4.00i + 1.00j)m/s at a point in the ocean where the position relative to a certain rock is r_i = (10.0i - 4.00j)m. After the fish swims with constant acceleration for 20.0s, its velocity v = (20.0i - 5.00j))m/s What are the components of the acceleration of the fish? What is the direction of its acceleration with respect to unit vector i If the fish maintains constant acceleration, where is it at t = 25.0 s and in what direction it is moving? An astronaut orbiting the earth is preparing to dock with a Westar VI Satellite. The satellite is in a circular orbit 600 km above the Earths surface, where the free fall acceleration is 8.21 m/s2. Take the radius of the earth as 6400 km, Determine the speed of the satellite and the time interval it requires to complete one orbit around the Earth. A science student is riding on a flatcar of train travelling along a straight, horizontal track at a constant speed of 10.0m/s. The student throws a ball into the air along a path that he judges to make an initial angle of 60.0 degree with the horizontal and to be in line with the track. The student's professor, who is standing on the ground nearby, observes the ball to rise vertically. How high does he see the ball rise?

Explanation / Answer

1)
Given
horizontal range of projectile is = 3*maximum height

       R = 3*H

   we know that the range of projectile is R = u^2 sin 2 theta/g

           max height H = u^2 sin^2 theta/2g

           u^2 sin 2 theta/g = 3(u^2 sin^2 theta/2g)

solving for theta , theta = 0 degrees

so the angle of projection is zero degrees

2)

given vectors

   V1 = 4i +1j m/s

   r = 10 i - 4 j m

   v2 = 20 i -5 j m/s

time t = 20 s

   acceleration ofthe fish is

   A = v2-v1/(t) = ((20 i -5 j)-(4i +1j))/20 = (1/20)(16i-4j)

   acceleratin vector is A = 0.8 i - 0.2 j

the direction is theta = arc tan (-0.2/0.8) ==> theta = -14.04 degrees

using equations of motion

   S = ut+0.5 at^2

after 25 s

   S = (4i +1j)25 + 0.5*(0.8 i - 0.2 j)25^2

   S = 100 i +25 j -0.5(500 i -125j)
   S = 100 i + 25j -250 i +62.5 j

   S = -150 i + 87.5 j

the direction is theta = arc tan (-87.5/150) = -30.26 degrees

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