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Math  /  Algebra

QuestionA ball is thrown upward at an initial velocity of 8.4 m/s8.4 \mathrm{~m} / \mathrm{s}, from a height of 1.5 m above the ground. The height of the ball, in metres, above the ground, after tt seconds, is modelled by the equation h=−4.9t2+8.4t+1.5h=-4.9 t^{2}+8.4 t+1.5. a) After how many seconds does the ball land on the ground? Round your answer to the nearest tenth of a second. b) What is the maximum height, to the nearest metre, that the ball reaches?

Studdy Solution

STEP 1

1. The motion of the ball is modeled by the quadratic equation h=−4.9t2+8.4t+1.5 h = -4.9t^2 + 8.4t + 1.5 .
2. The ball lands on the ground when its height h=0 h = 0 .
3. The maximum height is found at the vertex of the parabola described by the quadratic equation.

STEP 2

1. Solve for the time when the ball lands on the ground.
2. Determine the time at which the ball reaches its maximum height.
3. Calculate the maximum height using the time found in step 2.

STEP 3

Set the height equation to zero to find when the ball lands on the ground:
−4.9t2+8.4t+1.5=0 -4.9t^2 + 8.4t + 1.5 = 0

STEP 4

Solve the quadratic equation using the quadratic formula t=−b±b2−4ac2a t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} , where a=−4.9 a = -4.9 , b=8.4 b = 8.4 , and c=1.5 c = 1.5 .
t=−8.4±(8.4)2−4(−4.9)(1.5)2(−4.9) t = \frac{-8.4 \pm \sqrt{(8.4)^2 - 4(-4.9)(1.5)}}{2(-4.9)}

STEP 5

Calculate the discriminant b2−4ac b^2 - 4ac :
(8.4)2−4(−4.9)(1.5)=70.56+29.4=99.96 (8.4)^2 - 4(-4.9)(1.5) = 70.56 + 29.4 = 99.96

STEP 6

Calculate the values of t t :
t=−8.4±99.96−9.8 t = \frac{-8.4 \pm \sqrt{99.96}}{-9.8}
t=−8.4±9.998−9.8 t = \frac{-8.4 \pm 9.998}{-9.8}

STEP 7

Compute the two possible values for t t :
t1=−8.4+9.998−9.8≈−0.163 t_1 = \frac{-8.4 + 9.998}{-9.8} \approx -0.163
t2=−8.4−9.998−9.8≈1.98 t_2 = \frac{-8.4 - 9.998}{-9.8} \approx 1.98
Since time cannot be negative, we use t2 t_2 .

STEP 8

To find the time at which the ball reaches its maximum height, use the vertex formula t=−b2a t = -\frac{b}{2a} :
t=−8.42(−4.9)=8.49.8≈0.857 t = -\frac{8.4}{2(-4.9)} = \frac{8.4}{9.8} \approx 0.857

STEP 9

Substitute t=0.857 t = 0.857 into the height equation to find the maximum height:
h=−4.9(0.857)2+8.4(0.857)+1.5 h = -4.9(0.857)^2 + 8.4(0.857) + 1.5

STEP 10

Calculate the maximum height:
h≈−4.9(0.734)+8.4(0.857)+1.5 h \approx -4.9(0.734) + 8.4(0.857) + 1.5
h≈−3.5966+7.1988+1.5 h \approx -3.5966 + 7.1988 + 1.5
h≈5.1022 h \approx 5.1022
Round to the nearest metre:
h≈5 metres h \approx 5 \text{ metres}
The ball lands on the ground after approximately 2.0 2.0 seconds, and the maximum height reached is approximately 5 5 metres.

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