<span>Given:
Hmax (distance) = 50.0m
v</span>₀ = <span>70.0m/s
Required:
what angle should the arrow make with the horizontal as it is being shot
Solution:
Hmax = v</span>₀²sin²θ / 2g
sin²θ = 2gHmax / v₀²
sin²θ = 2 (9.81 m/s²) (50m) / (70 m/s)²
sin²θ = 0.200
θ = 26.56°
Answer:
<h2>
17.6 feet / second ^2</h2>
Explanation:
Explanation: 0 to 60 miles per hour means 0 to 88 feet/sec in five seconds (60 multiplied by 5280 (feet in one mile) divided by 3600 (seconds in an hour). The acceleration is 17.6 feet / second ^2.
Answer:
Explanation:
The initial velocity of the ball is
u = +25.9 m/s (towards the wall)
while the final velocity of the ball is
v = -21.3157 m/s (away from the wall)
The time taken for the change in velocity to occur is
t = 0.0133 s
The acceleration can be calculated as the change in velocity divided by the time taken:
Substituting the numbers, we find:
Answer:
12.495m/s
Explanation:
Horizontal displacement is the range of the projectile motion.
The range is expressed as;
R = 2U/g
U is the speed at which the rock is thrown (initial speed)
g is the acceleration due to gravity.
Given
R = 255cm = 2.55m
g = 9.8m/s²
Required
Speed U
Substitute the given parameters into the formula as shown;
2.55 = 2U/9.8
Cross multiply
2U = 2.55×9.8
2U = 24.99
U = 24.99/2
U = 12.495m/s
Hence the speed that you thew the rock is 12.495m/s
Answer:
Distance = 64 metres
Displacement = 28 metres
Direction = northward
Explanation:
Given that a kayaker moves 22 meters northward, then 18 meters southward , and finally 24 meters northward
Distance covered is only about the magnitude of the length covered.
Distance = 22 + 18 + 24 = 64 metres
The direction will be considered when calculating the displacement
Let northward be positive and southward be negative
Displacement = 22 - 18 + 24 = 28 metres
Since the displacement is positive, the direction of the motion is northward.
