Ask question

Ask question

All categories

3 years ago

9

A rookie quarterback throws a football with an initial upward velocity component of 17.0 m/s and a horizontal velocity component

of 18.3 m/s . Ignore air resistance. A. How much time is required for the football to reach the highest point of the trajectory?
B. How high is this point?
C. How much time (after it is thrown) is required for the football to return to its original level?
D. How does this compare with the time calculated in part (a).
E. How far has it traveled horizontally during this time?

1 answer:

vitfil [10]3 years ago

8 0

Answer:

(a) 1.73 s

(b) 14.75 m

(c) 3.36 s

(d) double

(e) 63.32 m

Explanation:

Vertical component of initial velocity, uy = 17 m/s

Horizontal component of initial velocity, ux = 18.3 m/s

(A) At highest point of trajectory, the vertical component of velocity is zero. Let the time taken is t.

Use first equation of motion in vertical direction

vy = uy - gt

0 = 17 - 9.8 t

t = 1.73 seconds

(B) Let the highest point is at height h.

Use III equation of motion in vertical direction

$v^{2}=u^{2}-2gh$

0 = 17 x 17 - 2 x 9.8 x h

h = 14.75 m

(C) The time taken by the ball to return to original level is T.

Use second equation of motion i vertical direction.

$h = ut + 0.5at^2$

h = 0 , u = 17 m/s

0 = 17 t - 0.5 x 9.8 t^2

t = 3.46 second

(D) It is the double of time calculated in part A

(E) Horizontal distance = horizontal velocity x total time

d = 18.3 x 3.46 = 63.32 m

You might be interested in

Define 1 kg of mass as international standard

Ann [662]

The kilogram is the Standard International System of Units unit of mass. It is defined as the mass of a particular international prototype made of platinum-iridium and kept at the International Bureau of Weights and Measures.

8 0

3 years ago

Read 2 more answers

As a stunt for movie two cars are to collide with each other head on. The two cars are initially 125 apart. Car A is heading str

Tema [17]

Answer:

3.39724 seconds

23.0824792352 m, 101.917520765 m

13.58896 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

The equation of motion will be

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow 125=30\times t+\dfrac{1}{2}\times 4\times t^2\\\Rightarrow 2t^2+30t-125=0\ m$

$t=\frac{5\left(\sqrt{19}-3\right)}{2},\:t=-\frac{5\left(3+\sqrt{19}\right)}{2}\\\Rightarrow t=3.39724, -18.39724$

The time at which the cars collide is 3.39724 seconds

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 4\times 3.39724^2\\\Rightarrow s=23.0824792352\ m$

Car B traveled 23.0824792352 m and Car A traveled 125-23.0824792352 = 101.917520765 m

$v=u+at\\\Rightarrow v=0+4\times 3.39724\\\Rightarrow v=13.58896\ m/s$

The speed of car B is 13.58896 m/s

4 0

3 years ago

Which diagram models the position of a soccer ball when it has the greatest amount of gravitational potential energy?

posledela

Answer:

C

Explanation:

Diagram C is the correct answer, because the ball is at the point with the highest height relative to the ground, in this way all the kinetic energy has been transformed into potential energy.

We must remember that potential energy is defined as the product of mass by gravity by height

Ep = m*g*h

where:

m = mass [kg]

g = gravity acceleration [m/s²]

h = elevation [m]

So when we have a great value for h in the above equation, we will have a big value for potential energy.

8 0

3 years ago

What is Acceleration due to gravity in cm/minute 2 ?

Xelga [282]

Answer:

f

Explanation:

:P

5 0

3 years ago

A 3kg ball moving at 8 m/s strikes a 2kg ball at rest,if the collision is elastic,what is the speed of the lighter ball if the h

zloy xaker [14]

Answer:

Speed of lighter ball is 4 m/s.

Explanation:

Applying the principle of conservation of linear momentum,

momentum before collision = momentum after collision.

$m_{1}$ $u_{1}$ + $m_{2}$ $u_{2}$ = $m_{1}$ $v_{1}$ - $m_{2}$ $v_{2}$

$m_{1}$ = 3 kg, $u_{1}$ = 8 m/s, $m_{2}$ = 2 kg, $u_{2}$ = 0 m/s ( since it is at rest), $v_{1}$ = 2 m/s, $v_{2}$ = ?

(3 x 8) + (2 x 0) = (8 x 2) - (2 x $v_{2}$ )

24 + 0 = 16 - 2 $v_{2}$

2 $v_{2}$ = 16 - 24

2 $v_{2}$ = -8

$v_{2}$ = $\frac{-8}{2}$

= -4 m/s

This implies that the light ball moves at the speed of 4 m/s in the opposite direction of the heavier ball after collision.

7 0

3 years ago