ball is thrown horizontally at 7m/s from the top of a hill that 42m high. how far from the base of the cliff would the ball hit the ground
how long does it take to fall 42m? (s = 1/2 at^2)
horizontal speed is constant, so distance = speed * time
h = Vo*T+gT^2 = 42
0+9.8T^2 = 42
T = 2.07s. to hit gnd.
d = V*T = 7*2.07 = 14.49m.
Correction: h = Vo*T+0.5gT^2 = 42
0+4.9T^2 = 42
T = 2.93s.
d = V*T = 7*2.93 = 20.5m.
To find how far the ball would hit the ground from the base of the cliff, we can use the kinematic equations of motion.
The given information is:
- Initial velocity of the ball (horizontal component): 7 m/s
- Height of the cliff: 42 m
First, we need to find the time it takes for the ball to hit the ground. Since the ball is thrown horizontally, the initial vertical velocity is 0 m/s. We can use the kinematic equation for vertical motion:
š = š£āš” + 1/2šš”Ā²
Where:
- š is the vertical displacement (ā42 m as the ball is falling downwards)
- š£ā is the initial vertical velocity (0 m/s)
- š is the acceleration due to gravity (ā9.8 m/sĀ² as it opposes the vertical motion and is downwards)
- š” is the time
Rearranging the equation gives:
ā42 = 0š” + 1/2(ā9.8)š”Ā²
ā42 = ā4.9š”Ā²
Now, solve for š” by rearranging the equation:
š”Ā² = 42/4.9
š” ā ā(8.57)
š” ā 2.93 s
Next, we need to find the horizontal distance traveled by the ball. The horizontal component of the initial velocity remains constant throughout the motion. We can use the equation:
š = š£š„š”
Where:
- š is the horizontal distance
- š£š„ is the horizontal component of the initial velocity (7 m/s)
- š” is the time (2.93 s)
Substituting the values:
š = 7 m/s Ć 2.93 s
š ā 20.5 m
Therefore, the ball would hit the ground approximately 20.5 meters from the base of the cliff.