A quarterback can throw a receiver a high, lazy \"lob\" pass or a low, quick \"b

Question

A quarterback can throw a receiver a high, lazy "lob" pass or a low, quick "bullet" pass. These passes are indicated by curves 1 and 2, respectively, in the figure (Figure 1).

Part A

The lob pass is thrown with an initial speed of 21.4 m/s and its time of flight is 3.98 s . What is its launch angle?

Express your answer using three significant figures.

Part B

The bullet pass is thrown with a launch angle of 25.0 . What is the initial speed of this pass?

Express your answer using two significant figures.

Part C

What is the time of flight of the bullet pass?

Express your answer using two significant figures.

1 of 1The graph shows height y as a function of distance x for trajectories of two passes. Distance is measured from 0 to 35 meters on the x axis. Height is measured from 0 to 20 meters on the y axis. Graph lines for both passes start from the origin and forming a symmetrical concave down curve land to 0 height at a distance of 35 meters. Curve 1 reaches the maximum height of approximately 19 meters, while curve 2 reaches the maximum height of approximately 4.8 meters.

A quarterback can throw a receiver a high, lazy "lob" pass or a low, quick "bullet" pass. These passes are indicated by curves 1 and 2, respectively, in the figure (Figure 1).

Part A

The lob pass is thrown with an initial speed of 21.4 m/s and its time of flight is 3.98 s . What is its launch angle?

Express your answer using three significant figures.

=     

Part B

The bullet pass is thrown with a launch angle of 25.0 . What is the initial speed of this pass?

Express your answer using two significant figures.

v0 =   m/s  

Part C

What is the time of flight of the bullet pass?

Express your answer using two significant figures.

t =   s   Figure

1 of 1The graph shows height y as a function of distance x for trajectories of two passes. Distance is measured from 0 to 35 meters on the x axis. Height is measured from 0 to 20 meters on the y axis. Graph lines for both passes start from the origin and forming a symmetrical concave down curve land to 0 height at a distance of 35 meters. Curve 1 reaches the maximum height of approximately 19 meters, while curve 2 reaches the maximum height of approximately 4.8 meters.

y (m) 15 10 x (m) 5 10 15 20 25 30 35

Explanation / Answer

a)

Considering motion along horizontal

X= u cos x*t

35 = 3.98 cos x* 21.4

x= 65.74

=========

b)

Range is given by

R= u^2 sin (2x)/g

35= u^2* sin (50)/9.8

u= 21.16 m/s

=======

C)

Time of flight

T= 2* u sin x/g

= 2*21.16* sin 25/9.8

= 1.825 second

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Comment in case any doubt.. good luck

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