1) 211m/s
2)240<span>°
3)759,600m or 759.6 km</span>
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:
brainly.com/question/24136952
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The longitude based on the time difference is 15 degrees.
<h3>Longitude of complete rotation of the Earth</h3>
The longitude of a complete rotation of the earth in a 24 hours is calculated as follows;
<h3>Time difference</h3>
The time difference between the local apparent solar time and the Greenwich time is calculated as follows;
Since it is one hour time difference, the longitude is 15 degrees.
Learn more about Earth longitude here: brainly.com/question/1939015
Explanation:
When hot water is poured on the can in a bucket of cold water, the can crushes off means it gets unshaped
