Answer:
Time = 80.91 seconds
Explanation:
Given the following data;
Velocity = 5.50 m/s.
Distance = 445 meters
To find the time;
Velocity can be defined as the rate of change in displacement (distance) with time. Velocity is a vector quantity and as such it has both magnitude and direction.
Mathematically, velocity is given by the equation;

Substituting into the formula, we have;
5.5 = 445/time
Time = 445/5.5
Time = 80.91 seconds
Answer: 2940 J
Explanation: solution attached:
PE= mgh
Substitute the values:
PE= 10kg x 9.8 m/s² x 30 m
= 2940 J
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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