with an initial speed of 8.0 m/s at an angle of 40 degrees above the horizontal. The ball leaves her hand 1.0 meters above the ground. See full question below.

Question

A child throws a ball with an initial speed of 8.0 m/s at an angle of 40 degrees above the horizontal. The ball leaves her hand 1.0 meters above the ground. At what angle below the horizontal does the ball approach the ground?

The answer options are:

A) 35 degrees

B) 42 degrees

C) 48 degrees

D) 40 degrees

E) 65 degrees

 

I would like to know how to solve this question, rather than just know the answer, as I may encounter such questions in the future.

Answer 1

A child throws a ball with an initial speed of 8.0 m/s at an angle of 40 degrees above the horizontal. The ball leaves her hand 1.0 meters above the ground. At what angle below the horizontal does the ball approach the ground?
The answer options are:
A) 35 degrees
B) 42 degrees
C) 48 degrees
D) 40 degrees
E) 65 degrees

 

===================================

 

Imagine a 1 meter high (from ground) level.

Consider the moment in time when the ball reaches this level, after descending from the peak of trajectory.

So, the ball at this instant has still 8 m/sec speed, and it moves at 40 degree BELOW our level (to the ground).

Horizontal component of velocity would be

 

Vx=8*cos(40)=6.128 (m/sec)

 

The vertical component will be same as in initial state, only with opposite sign - it goes down:

 

Vy=8*sin(40)= 5.1423 (m/sec)

 

The problem transforms to the following scenario:

The ball falls with these velocities (Vx and Vy) in gravitational field of earth obviously.

We need to find final components of velocity of the ball when it hits the ground.

 

Let them be Vxt and Vyt.

 

Then, the angle, below horizontal,  at which ball approach the ground will be,

 

Theta - atan( Vyt/Vxt)

 

Notice, that Vxt=Vx= 6.128 (m/sec) [ it's constant along the entire trajectory ...]

 

Vyt can be found form usual connection:

 

Vyt = Vy + g*t

 

But, we don't have the time - t

It can be found from ;

 

H = Vyt+g*t^2/2  ; H=1 meter inour case ...

 

So, we end up with quadratic equation for t:

 

gt^2/2+Vyt-H=0 , or

 

4.9*t^2+5.142*t-1=0

 

It has two solutions - one of them (the negative one) isn't physical.

Take the positive solution:

 

t= 0.168 sec

 

Then

 

Theta = atan ( 6.788/6.129)=47.93 degrees

 

This must be the choice C.

 

I'm leaving (for you) room for double checking the arithmetic. 

I hope u got the idea ...

message if still questions:

 

Good luck, and

Happy New Year :)