A ball is thrown upward from the top of a 35-m tower, with an initial velocity of 80 m/s and an angle of 25 degree. a) Find the time to reach the ground and the distance R from the bottom of the tower to the point of impact in the figure below. b) Find the magnitude and direction of the velocity at the moment of impact.
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Expert's answer
2021-01-12T12:10:29-0500
As per the given question,
Height of the tower (h)=35m
Initial velocity of the ball (v) = 80 m/s
Angle of projection (θ)=25∘
a) Time to reach the ground (t)=?
Distance from the bottom of the tower to the point of projection (R)=?
b) Magnitude and the direction of velocity, at the time of impact to the ground=?
Let the final velocity of the ball be v1
Now, taking the vertical and horizontal component of initial velocity.
Vertical component of velocity along y axis,(vy)=vsin25∘
Velocity along the x axis (vx)=vcos25∘
Now, applying the second law of motion,
−h=ut−2gt2
Now, substituting the values in the above equation,
⇒−35=vsin25∘t−2gt2
⇒−35=80sin25∘t−29.8t2
⇒4.9t2−33.8t−35=0
comparing it with standard quadratic equation ax2+bx+c=0
a=4.9, b =-33.8, c = -35
⇒t=2a−b±b2−4ac
Now, substituting the values in the above,
⇒t=2×4.9−(−33.8)±(−33.8)2+4×4.9×(−35)
⇒t=9.833.8±1828.44s
⇒t=9.833.8±42.76s
time can not be negative, hence we will consider positive value,
⇒t=9.833.8+42.76s
⇒t=9.876.56s
⇒t=7.81s
Range of the ball (R)=vcos25∘×t
Now, substituting the values,
R=80cos25∘×7.81m
⇒R=566.26m from the the bottom of the tower to the point of impact.
b) Horizontal component of the velocity remain unchanged, because there is not any force in the horizontal direction, but the vertical component of the velocity will not constant, because there is gravity force in the vertical direction.
At the point of the impact to the ground, let the angle made by the ball is ϕ
Hence v1sinϕ=vsin25∘−gt
now substituting the values
⇒v1sinϕ=80sin25∘−9.8×7.81
⇒v1sinϕ=−42.72....(i)
the horizontal component velocity remain unchanged hence,
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