Because mass does not change from place to place but weight does change from place to place... why? because weight is the amount of gravitational force on an object and mass is the amount of matter in an object. mars has less gravitational force so an object will weigh less than it really weighs there
        
             
        
        
        
Answer:
a)    x = v₀² sin 2θ / g
b)    t_total = 2 v₀ sin θ / g
c)    x = 16.7 m
Explanation:
This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity
         sin θ =  / vo
 / vo
         cos θ = v₀ₓ / vo
          v_{oy} = v_{o} sin θ
          v₀ₓ = v₀ cos θ
          v_{oy} = 13.5 sin 32 = 7.15 m / s
          v₀ₓ = 13.5 cos 32 = 11.45 m / s
a) In the x axis there is no acceleration so the velocity is constant
          v₀ₓ = x / t
           x = v₀ₓ t
the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero
           v_{y} = v_{oy} - gt
           0 = v₀ sin θ - gt
           t = v_{o} sin θ / g
          
we substitute
        x = v₀ cos θ (2 v_{o} sin θ / g)
        x = v₀² /g      2 cos θ sin θ
        x = v₀² sin 2θ / g
at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,
b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time
at the highest point the vertical speed is zero
           v_{y} = v_{oy} - gt
           v_{y} = 0
            t = v_{oy} / g
            t = v₀ sin θ / g
as the time to get on and off is the same the total time or flight time is
            t_total = 2 t
            t_total = 2 v₀ sin θ / g
c) we calculate
           x = 13.5 2 sin (2 32) / 9.8
           x = 16.7 m
 
        
             
        
        
        
Answer:
v₀ = 16.55 m/s
Explanation:
This motion of the ball can be modeled as a projectile motion with following data:
R = Range of Projectile = 27.5 m
θ = Launch Angle = 50°
g = acceleration due to gravity = 9.81 m/s²
v₀ = Initial Speed of Ball = ?
Therefore, using formula for range of projectile, we have:

<u>v₀ = 16.55 m/s</u>
 
        
             
        
        
        
Explanation:
Given that,
Rate of cooling of air 
Initial temperature= 80°C
Final temperature = 5°C
We need to calculate 
Using newton's law of cooling


Where, 
Here,  
 
 = 25°C  (surrounding temperature)
 = 25°C  (surrounding temperature)
dt = 1 minute

Put the value into the formula



Hence, This is the required answer.