Question

1. A soccer ball is kicked at an upward angle θ between 0° and 90°. Its initial vertical (y) velocity vy0 is 19.6 m/s. It follows a parabolic trajectory and lands at the end of its range.

c. If the ball’s range is 60 m, is the angle θ less than, equal to, or greater than 45°?

2. A small toy car is placed upon a ramp that is 2 m long. The car is released (with zero initial velocity) from the top of the ramp. Assume that friction is negligible.(aramp= 1 m/s2)

How high is the top if the car takes 2.0 s to reach the bottom?

Answer 1

#1

As the ball is projected at some angle with vertical speed

`v_(y0) = 19.6 m/s`

now we can find the time of flight

`T = (2v_(y0))/(g)`

`T = (2*19.6)/(9.8)`

`T = 4 s`

now we can find the horizontal speed as

Range = horizontal speed * time

`60 = v_(x0)*T`

`60 = v_(x0)* 4`

`v_(x0) = 15 m/s`

now for the projection angle we will have

`tan\theta = (v_(y0))/(v_(x0))`

`tan\theta = (19.6)/(15)`

so here angle is more than 45 degree as the above ratio is more than 1

#2

here we know that car takes t = 2 s to reach the bottom

here acceleration = 1 m/s^2

initial speed = 0

we can use kinematics now

`d = v_i*t + (1)/(2) at^2`

`d = 0 + (1)/(2)*1*2^2`

`d = 2 m`

so the toy car is released from the top at distance 2 m from the bottom