Solution Found!
A projectile is shot from the edge of a cliff 125 m above ground level with an initial
Chapter 3, Problem 31P(choose chapter or problem)
(II) A projectile is shot from the edge of a cliff 125 m above the ground level with an initial speed of 65.0 m/s at an angle of \(37.0^{\circ}\) with the horizontal, as shown in Fig. 3-35. (a) Determine the time taken by the projectile to hit point \(P\) at ground level. (b) Determine the range \(X\) of the projectile as measured from the base of the cliff. At the instant just before the projectile hits point \(P\), find (c) the horizontal and the vertical components of its velocity, (d) the magnitude of the velocity, and (e) the angle made by the velocity vector with the horizontal. (f) Find the maximum height above the cliff top reached by the projectile.
Figure 3-35
Equation Transcription:
Text Transcription:
37.0 degrees
P
X
Questions & Answers
QUESTION:
(II) A projectile is shot from the edge of a cliff 125 m above the ground level with an initial speed of 65.0 m/s at an angle of \(37.0^{\circ}\) with the horizontal, as shown in Fig. 3-35. (a) Determine the time taken by the projectile to hit point \(P\) at ground level. (b) Determine the range \(X\) of the projectile as measured from the base of the cliff. At the instant just before the projectile hits point \(P\), find (c) the horizontal and the vertical components of its velocity, (d) the magnitude of the velocity, and (e) the angle made by the velocity vector with the horizontal. (f) Find the maximum height above the cliff top reached by the projectile.
Figure 3-35
Equation Transcription:
Text Transcription:
37.0 degrees
P
X
ANSWER:
Step 1 of 6
a.)
We have to determine the time taken by the projectile to hit point 
The time taken by the projectile to reach the ground can be found using the equation,
Upward 
Now,
direction =
= 
Thus,
=
=
= 
Hence, 
Therefore, the time taken by the projectile to reach the ground is 10.4 s.