A shot-putter released the shot at an angle of 41.5 degrees and a height of 1.9 m with an initial velocity of 13.3 m/s. How far
1 answer:
Answer:
x = 17.88[m]
Explanation:
We can find the components of the initial velocity:
We have to remember that the acceleration of gravity will be worked with negative sign, since it acts in the opposite direction to the movement in direction and the projectile upwards.
g = - 9.81[m/s^2]
Now we must find the time it takes for the projectile to hit the ground, as the problem mentions that it does not impact on the board.
With this time we can calculate the horizontal distance:
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Answer:
2(maximum), -2(minimum), -2(maximum).
Explanation:
V(t)= 2πcos πt--------------------------------------------------------------------------------(1).
Therefore, there is a need to integrate v(t) to get S(t).
S(t)= 2×sinπt + C ------------------------------------------------------------------------------(2).
Applying the condition given, we have s(0)= 0.
S(0)= 2sin ×π(0) + C.
Which means that; 0+C= 0. That is; C=0.
S(t)= 2 sin πt.
The mass moves to its highest positions at time,t=half(1/2=.5) and time,t=2.5.
Take note that; sin(π/2) = sin(5π/2) = 1 .
Also, the mass moves to its lowest position at time,t=(3/2); also, sin(3π/2) = -1.
Therefore, we have that 2 maximum; -2 minimum and -2 maximum.
