The frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.
To find the answer, we need to know about the time of flight and range of projectile motion.
<h3>What's the expression of range of a projectile motion?</h3>
- Range = U²× sin(2θ)/g
- U= initial velocity, θ= angle of projectile and g= acceleration due to gravity
- U=√{Range×g/sin(2θ)}
- Here, range= 2.20m, = 36.5°
- U= √{2.20×9.8/sin(73)}
U= √{2.20×9.8/sin(73)} = 22.5m/s
<h3>What's the expression of time of flight in projectile motion?</h3>
- Time of flight= (2×U×sinθ)/g
- So, T= (2×22.5×sin36.5°)/9.8
= 2.73 s
Thus, we can conclude that the frog's launch speed and the time spends in the air are 22.5m/s and 2.73s respectively.
Learn more about the range and time period of projectile motion here:
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Answer:
a. slope=rise/run
rise=0.02
run=-2
determined using the point (3,0.08) and (1,0.1) on the graph
slope=0.02/-2
= -0.01 or -1/100
b.area= area of trapizoid+ rectangle
((0.07+0.11)÷2)×4+1×0.07
0.36+0.07
=0.43$
c. the area represent the total cost after 5 hours
PLEASE MARK BRAINLIEST
Answer:
3.0 seconds
Explanation:
We can solve the problem by considering the horizontal motion of the ball only. In fact, the ball moves by uniform motion (constant speed) along the horizontal direction, since there are no forces acting in this direction. The horizontal speed of the ball is given by:
and it does not change during the motion.
We also know that the ball travels a horizontal distance of d = 60 m, so we can find the time it takes to cover the distance by using the equation:
Answer:
5 miles a second
Explanation:
20 divided by 4
hope it helps and for brainliest :)
