(I) A tiger leaps horizontally from a 7.5-m-high rock with a speed of 3.2 m/s. How far from the base of the rock will she land? Choose downward to be the positive y direction. The origin will be at the point where the tiger leaps from the rock. In the horizontal direction. v x 0 = 3.2 m/s and a x = 0. In the vertical direction, v y 0 = 0, a y = 9.80 m/s 2 , y 0 = 0 , and the final location y = 7.5 m. The time for the tiger to reach the ground is found from applying Eq. 2-12b to the vertical motion. y = y 0 + v y 0 t + 1 2 a y t 2 → 7.5 m = 0 + 0 + 1 2 ( 9.80 m/s 2 ) t 2 → t = 2 ( 7.5 m ) 9.80 m / s 2 = 1.24 sec The horizontal displacement is calculated from the constant horizontal velocity. Δ x = v x t = (3.2 m/s) (1.24 sec) = 4.0 m
(I) A tiger leaps horizontally from a 7.5-m-high rock with a speed of 3.2 m/s. How far from the base of the rock will she land? Choose downward to be the positive y direction. The origin will be at the point where the tiger leaps from the rock. In the horizontal direction. v x 0 = 3.2 m/s and a x = 0. In the vertical direction, v y 0 = 0, a y = 9.80 m/s 2 , y 0 = 0 , and the final location y = 7.5 m. The time for the tiger to reach the ground is found from applying Eq. 2-12b to the vertical motion. y = y 0 + v y 0 t + 1 2 a y t 2 → 7.5 m = 0 + 0 + 1 2 ( 9.80 m/s 2 ) t 2 → t = 2 ( 7.5 m ) 9.80 m / s 2 = 1.24 sec The horizontal displacement is calculated from the constant horizontal velocity. Δ x = v x t = (3.2 m/s) (1.24 sec) = 4.0 m
(I) A tiger leaps horizontally from a 7.5-m-high rock with a speed of 3.2 m/s. How far from the base of the rock will she land?
Choose downward to be the positive y direction. The origin will be at the point where the tiger leaps from the rock. In the horizontal direction. vx0 = 3.2 m/s and ax = 0. In the vertical direction, vy0 = 0, ay = 9.80 m/s2, y0 = 0, and the final location y = 7.5 m. The time for the tiger to reach the ground is found from applying Eq. 2-12b to the vertical motion.
y
=
y
0
+
v
y
0
t
+
1
2
a
y
t
2
→
7.5
m
=
0
+
0
+
1
2
(
9.80
m/s
2
)
t
2
→
t
=
2
(
7.5
m
)
9.80
m
/
s
2
=
1.24
sec
The horizontal displacement is calculated from the constant horizontal velocity.
A basketball player is standing on the floor 14.0 m away from the basket. The height of the basket is 3.05 m, and heshoots the ball at 49.0° with the horizontal from a height of 2.00 m. At what speed must the player shoot the ball so thatit goes through the hoop without striking the backboard?
If an arrow is shot straight upward on the surface of the moon with initialvelocity 60m/s, its height after t seconds is given byh(t) = 60t − 0.8t^2.
What is the velocity of the arrow when it hits the moon?
Question 2:
#63 In 1780, in what is now referred to as "Brady's Leap," Captain Sam Brady of the U.S. Continental Army escaped certain death from his enemies by running horizontally off the edge of the cliff above Ohio's Cuyahoga River, which is con-fined at that spot to a gorge. He landed safely on the far side of the river. It was reported that he leapt 22 ft across while falling 20 ft. Tall tale, or possible?
Chapter 3 Solutions
Physics for Scientists and Engineers with Modern Physics
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.